GSEB Class 12 Maths Solutions Chapter 9 Differential Equations Exercise 9.5

In each of the questions 1 to 10, show that the given differential equation is homogeneous and solve each of them:

Question 1. \( (x^2 + xy) dy = (x^2 + y^2)dx \)
Answer: First, we can rewrite the given differential equation in the form \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy} = f(x, y) \) (let's call this equation (1))
To show that \( f(x, y) \) is a homogeneous function, we replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{(\lambda x)^2 + (\lambda y)^2}{(\lambda x)^2 + (\lambda x)(\lambda y)} \)
\( = \frac{\lambda^2 x^2 + \lambda^2 y^2}{\lambda^2 x^2 + \lambda^2 xy} \)
\( = \frac{\lambda^2 (x^2 + y^2)}{\lambda^2 (x^2 + xy)} \)
\( = \lambda^0 \left(\frac{x^2 + y^2}{x^2 + xy}\right) \)
\( = \lambda^0 f(x, y) \)
Since \( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \), the function \( f(x, y) \) is a homogeneous function of degree zero.
Now, to solve the differential equation, we make the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into equation (1):
\( v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{x^2 + x(vx)} \)
\( v + x \frac{dv}{dx} = \frac{x^2 + v^2 x^2}{x^2 + vx^2} \)
\( v + x \frac{dv}{dx} = \frac{x^2(1 + v^2)}{x^2(1 + v)} \)
\( v + x \frac{dv}{dx} = \frac{1 + v^2}{1 + v} \)
Now, we separate the variables:
\( x \frac{dv}{dx} = \frac{1 + v^2}{1 + v} - v \)
\( x \frac{dv}{dx} = \frac{1 + v^2 - v(1 + v)}{1 + v} \)
\( x \frac{dv}{dx} = \frac{1 + v^2 - v - v^2}{1 + v} \)
\( x \frac{dv}{dx} = \frac{1 - v}{1 + v} \)
\( \frac{1 + v}{1 - v} dv = \frac{dx}{x} \)
Now, integrate both sides:
\( \int \frac{1 + v}{1 - v} dv = \int \frac{dx}{x} \)
We can rewrite the left side as \( \int \frac{2 - (1 - v)}{1 - v} dv = \int \left(\frac{2}{1 - v} - 1\right) dv \):
\( \int \left(\frac{2}{1 - v} - 1\right) dv = \int \frac{dx}{x} \)
\( 2 \int \frac{1}{1 - v} dv - \int 1 dv = \int \frac{1}{x} dx \)
\( 2(-\log|1 - v|) - v = \log|x| + \log C \)
\( -2 \log|1 - v| - v = \log|x| + \log C \)
Now, substitute back \( v = \frac{y}{x} \):
\( -2 \log\left|1 - \frac{y}{x}\right| - \frac{y}{x} = \log|x| + \log C \)
\( -2 \log\left|\frac{x - y}{x}\right| - \frac{y}{x} = \log|x| + \log C \)
\( -2 (\log|x - y| - \log|x|) - \frac{y}{x} = \log|x| + \log C \)
\( -2 \log|x - y| + 2 \log|x| - \frac{y}{x} = \log|x| + \log C \)
Rearrange the terms:
\( \log|x| - 2 \log|x - y| - \frac{y}{x} = \log C \)
\( \log\left|\frac{x}{(x - y)^2}\right| - \frac{y}{x} = \log C \)
\( \log\left|\frac{x}{(x - y)^2}\right| = \frac{y}{x} + \log C \)
To eliminate the logarithm, we use exponential function:
\( \frac{x}{(x - y)^2} = e^{\frac{y}{x} + \log C} \)
\( \frac{x}{(x - y)^2} = e^{\frac{y}{x}} e^{\log C} \)
Let \( e^{\log C} = K \) (a new constant).
\( \frac{x}{(x - y)^2} = K e^{\frac{y}{x}} \)
\( (x - y)^2 = \frac{x}{K} e^{-\frac{y}{x}} \)
Let \( \frac{1}{K} = C_1 \) (another constant).
\( (x - y)^2 = C_1 x e^{-\frac{y}{x}} \)
This is the general solution.

In simple words: First, we checked if the equation was "homogeneous" by substituting \( \lambda x \) and \( \lambda y \). Since it was, we used a special trick: we replaced \( y \) with \( vx \) and \( \frac{dy}{dx} \) with \( v + x \frac{dv}{dx} \). This allowed us to separate the parts with \( v \) and \( x \) and then integrate both sides. Finally, we put \( \frac{y}{x} \) back for \( v \) to get the final answer.

Exam Tip: Remember to always first check for homogeneity. If it's homogeneous, the substitution \( y=vx \) is key, as it transforms the differential equation into a separable form that is easier to integrate.

Question 2. \( y' = \frac{x+y}{x} \)
Answer: We have the given differential equation \( y' = \frac{x+y}{x} \).
This can be written as \( \frac{dy}{dx} = \frac{x+y}{x} = f(x, y) \).
To show that \( f(x, y) \) is a homogeneous function, we replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x} \)
\( = \frac{\lambda(x+y)}{\lambda x} \)
\( = \lambda^0 \left(\frac{x+y}{x}\right) \)
\( = \lambda^0 f(x, y) \)
Since \( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \), the function \( f(x, y) \) is a homogeneous function of degree zero.
To solve the differential equation, we make the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into the equation:
\( v + x \frac{dv}{dx} = \frac{x + vx}{x} \)
\( v + x \frac{dv}{dx} = \frac{x(1 + v)}{x} \)
\( v + x \frac{dv}{dx} = 1 + v \)
Now, separate the variables:
\( x \frac{dv}{dx} = 1 + v - v \)
\( x \frac{dv}{dx} = 1 \)
\( dv = \frac{dx}{x} \)
Now, integrate both sides:
\( \int dv = \int \frac{dx}{x} \)
\( v = \log|x| + C \)
Substitute back \( v = \frac{y}{x} \):
\( \frac{y}{x} = \log|x| + C \)
Multiply by \( x \) to get the final solution:
\( y = x \log|x| + Cx \)

In simple words: We first checked if the equation was homogeneous, which it was. Then, we replaced \( y \) with \( vx \) and \( y' \) with \( v + x \frac{dv}{dx} \). After simplifying, we separated the \( v \) and \( x \) terms and integrated them. Finally, we put \( \frac{y}{x} \) back for \( v \) to find the answer.

Exam Tip: For simple homogeneous equations like this, the substitution \( y=vx \) often leads to a direct integration. Be careful with constant of integration, often denoted as \( C \).

Question 3. \( (x - y)dy - (x + y)dx = 0 \)
Answer: We have the given differential equation \( (x - y)dy - (x + y)dx = 0 \).
We can rewrite this in the form \( \frac{dy}{dx} \):
\( (x - y)dy = (x + y)dx \)
\( \frac{dy}{dx} = \frac{x + y}{x - y} = f(x, y) \)
To show that \( f(x, y) \) is a homogeneous function, we replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x - \lambda y} \)
\( = \frac{\lambda(x + y)}{\lambda(x - y)} \)
\( = \lambda^0 \left(\frac{x + y}{x - y}\right) \)
\( = \lambda^0 f(x, y) \)
Since \( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \), the function \( f(x, y) \) is a homogeneous function of degree zero.
To solve the differential equation, we make the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into the equation:
\( v + x \frac{dv}{dx} = \frac{x + vx}{x - vx} \)
\( v + x \frac{dv}{dx} = \frac{x(1 + v)}{x(1 - v)} \)
\( v + x \frac{dv}{dx} = \frac{1 + v}{1 - v} \)
Now, separate the variables:
\( x \frac{dv}{dx} = \frac{1 + v}{1 - v} - v \)
\( x \frac{dv}{dx} = \frac{1 + v - v(1 - v)}{1 - v} \)
\( x \frac{dv}{dx} = \frac{1 + v - v + v^2}{1 - v} \)
\( x \frac{dv}{dx} = \frac{1 + v^2}{1 - v} \)
\( \frac{1 - v}{1 + v^2} dv = \frac{dx}{x} \)
Now, integrate both sides:
\( \int \frac{1 - v}{1 + v^2} dv = \int \frac{dx}{x} \)
We can split the integral on the left side:
\( \int \frac{1}{1 + v^2} dv - \int \frac{v}{1 + v^2} dv = \int \frac{1}{x} dx \)
\( \tan^{-1} v - \frac{1}{2} \int \frac{2v}{1 + v^2} dv = \int \frac{1}{x} dx \)
\( \tan^{-1} v - \frac{1}{2} \log|1 + v^2| = \log|x| + C \)
Substitute back \( v = \frac{y}{x} \):
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log\left|1 + \left(\frac{y}{x}\right)^2\right| = \log|x| + C \)
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log\left|\frac{x^2 + y^2}{x^2}\right| = \log|x| + C \)
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} (\log|x^2 + y^2| - \log|x^2|) = \log|x| + C \)
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log|x^2 + y^2| + \frac{1}{2} \log|x^2| = \log|x| + C \)
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log(x^2 + y^2) + \frac{1}{2} (2 \log|x|) = \log|x| + C \)
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log(x^2 + y^2) + \log|x| = \log|x| + C \)
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log(x^2 + y^2) = C \)
This is the general solution.

In simple words: We changed the given equation into the standard \( \frac{dy}{dx} \) form. We confirmed it was homogeneous by using \( \lambda x \) and \( \lambda y \). Then, we substituted \( y = vx \) to simplify, separated the \( v \) and \( x \) terms, and integrated each side. Finally, we put \( \frac{y}{x} \) back for \( v \) to get the solution.

Exam Tip: When integrating expressions involving \( \frac{v}{1+v^2} \), remember the form \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C \). Also, carefully handle the properties of logarithms during simplification, especially when dealing with \( \log|x^2| \).

Question 4. \( (x^2 - y^2)dx + 2xy dy = 0 \)
Answer: We have the given differential equation \( (x^2 - y^2)dx + 2xy dy = 0 \).
We can rewrite this in the form \( \frac{dy}{dx} \):
\( 2xy dy = -(x^2 - y^2)dx \)
\( 2xy dy = (y^2 - x^2)dx \)
\( \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} = f(x, y) \)
To show that \( f(x, y) \) is a homogeneous function, we replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{(\lambda y)^2 - (\lambda x)^2}{2(\lambda x)(\lambda y)} \)
\( = \frac{\lambda^2 y^2 - \lambda^2 x^2}{2\lambda^2 xy} \)
\( = \frac{\lambda^2 (y^2 - x^2)}{\lambda^2 (2xy)} \)
\( = \lambda^0 \left(\frac{y^2 - x^2}{2xy}\right) \)
\( = \lambda^0 f(x, y) \)
Since \( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \), the function \( f(x, y) \) is a homogeneous function of degree zero.
To solve the differential equation, we make the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into the equation:
\( v + x \frac{dv}{dx} = \frac{(vx)^2 - x^2}{2x(vx)} \)
\( v + x \frac{dv}{dx} = \frac{v^2 x^2 - x^2}{2vx^2} \)
\( v + x \frac{dv}{dx} = \frac{x^2(v^2 - 1)}{x^2(2v)} \)
\( v + x \frac{dv}{dx} = \frac{v^2 - 1}{2v} \)
Now, separate the variables:
\( x \frac{dv}{dx} = \frac{v^2 - 1}{2v} - v \)
\( x \frac{dv}{dx} = \frac{v^2 - 1 - 2v^2}{2v} \)
\( x \frac{dv}{dx} = \frac{-v^2 - 1}{2v} \)
\( x \frac{dv}{dx} = -\frac{v^2 + 1}{2v} \)
\( \frac{2v}{v^2 + 1} dv = -\frac{dx}{x} \)
Now, integrate both sides:
\( \int \frac{2v}{v^2 + 1} dv = -\int \frac{1}{x} dx \)
\( \log|v^2 + 1| = -\log|x| + \log C \)
\( \log(v^2 + 1) = \log\left|\frac{C}{x}\right| \)
\( v^2 + 1 = \frac{C}{x} \)
Substitute back \( v = \frac{y}{x} \):
\( \left(\frac{y}{x}\right)^2 + 1 = \frac{C}{x} \)
\( \frac{y^2}{x^2} + 1 = \frac{C}{x} \)
\( \frac{y^2 + x^2}{x^2} = \frac{C}{x} \)
Multiply by \( x^2 \):
\( y^2 + x^2 = Cx \)
This is the general solution.

In simple words: First, we rearranged the equation to get \( \frac{dy}{dx} \). We checked that it was homogeneous by substituting \( \lambda x \) and \( \lambda y \). Then, we used the substitution \( y = vx \) to make the equation simpler. After separating the terms with \( v \) and \( x \), we integrated both sides. Finally, we replaced \( v \) with \( \frac{y}{x} \) to find the overall answer.

Exam Tip: Pay close attention to the algebraic manipulation when separating variables, especially sign changes. The integral \( \int \frac{2v}{v^2+1} dv \) is a common pattern that results in a logarithm, so recognize this quickly.

Question 5. \( x^2 \frac{dy}{dx} = x^2 - 2y^2 + xy \)
Answer: We have the given differential equation \( x^2 \frac{dy}{dx} = x^2 - 2y^2 + xy \).
We can rewrite this in the form \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{x^2 - 2y^2 + xy}{x^2} = f(x, y) \)
To show that \( f(x, y) \) is a homogeneous function, we replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{(\lambda x)^2 - 2(\lambda y)^2 + (\lambda x)(\lambda y)}{(\lambda x)^2} \)
\( = \frac{\lambda^2 x^2 - 2\lambda^2 y^2 + \lambda^2 xy}{\lambda^2 x^2} \)
\( = \frac{\lambda^2 (x^2 - 2y^2 + xy)}{\lambda^2 x^2} \)
\( = \lambda^0 \left(\frac{x^2 - 2y^2 + xy}{x^2}\right) \)
\( = \lambda^0 f(x, y) \)
Since \( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \), the function \( f(x, y) \) is a homogeneous function of degree zero.
To solve the differential equation, we make the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into the equation:
\( v + x \frac{dv}{dx} = \frac{x^2 - 2(vx)^2 + x(vx)}{x^2} \)
\( v + x \frac{dv}{dx} = \frac{x^2 - 2v^2 x^2 + vx^2}{x^2} \)
\( v + x \frac{dv}{dx} = \frac{x^2(1 - 2v^2 + v)}{x^2} \)
\( v + x \frac{dv}{dx} = 1 - 2v^2 + v \)
Now, separate the variables:
\( x \frac{dv}{dx} = 1 - 2v^2 + v - v \)
\( x \frac{dv}{dx} = 1 - 2v^2 \)
\( \frac{dv}{1 - 2v^2} = \frac{dx}{x} \)
Now, integrate both sides:
\( \int \frac{dv}{1 - 2v^2} = \int \frac{dx}{x} \)
We can factor out \( 2 \) from the denominator:
\( \int \frac{dv}{2\left(\frac{1}{2} - v^2\right)} = \int \frac{dx}{x} \)
\( \frac{1}{2} \int \frac{dv}{\left(\frac{1}{\sqrt{2}}\right)^2 - v^2} = \int \frac{dx}{x} \)
Using the standard integral formula \( \int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \log\left|\frac{a + x}{a - x}\right| \), where \( a = \frac{1}{\sqrt{2}} \):
\( \frac{1}{2} \left[\frac{1}{2 \cdot \frac{1}{\sqrt{2}}} \log\left|\frac{\frac{1}{\sqrt{2}} + v}{\frac{1}{\sqrt{2}} - v}\right|\right] = \log|x| + \log C \)
\( \frac{1}{2} \left[\frac{\sqrt{2}}{2} \log\left|\frac{1 + \sqrt{2}v}{1 - \sqrt{2}v}\right|\right] = \log|x| + \log C \)
\( \frac{\sqrt{2}}{4} \log\left|\frac{1 + \sqrt{2}v}{1 - \sqrt{2}v}\right| = \log|x| + \log C \)
Substitute back \( v = \frac{y}{x} \):
\( \frac{\sqrt{2}}{4} \log\left|\frac{1 + \sqrt{2}\frac{y}{x}}{1 - \sqrt{2}\frac{y}{x}}\right| = \log|x| + \log C \)
\( \frac{\sqrt{2}}{4} \log\left|\frac{x + \sqrt{2}y}{x - \sqrt{2}y}\right| = \log|x| + \log C \)
This is the general solution.

In simple words: First, we put the equation in \( \frac{dy}{dx} \) form and verified it was homogeneous. Next, we used the substitution \( y = vx \) to make the equation separable. We then split the \( v \) and \( x \) terms and integrated them. Finally, we substituted \( \frac{y}{x} \) back for \( v \) to complete the solution.

Exam Tip: Be mindful of using the correct integral formulas, especially for inverse trigonometric or logarithmic forms. Also, keep track of constants and simplify the final answer by substituting back \( v = \frac{y}{x} \).

Question 6. \( x dy - y dx = \sqrt{x^2+y^2} dx \)
Answer: We have the given differential equation \( x dy - y dx = \sqrt{x^2+y^2} dx \).
Rearrange the terms to get \( \frac{dy}{dx} \):
\( x dy = y dx + \sqrt{x^2+y^2} dx \)
\( x dy = (y + \sqrt{x^2+y^2}) dx \)
\( \frac{dy}{dx} = \frac{y + \sqrt{x^2+y^2}}{x} = f(x, y) \)
To show that \( f(x, y) \) is a homogeneous function, we replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{(\lambda x)^2+(\lambda y)^2}}{\lambda x} \)
\( = \frac{\lambda y + \sqrt{\lambda^2 x^2+\lambda^2 y^2}}{\lambda x} \)
\( = \frac{\lambda y + \lambda \sqrt{x^2+y^2}}{\lambda x} \)
\( = \frac{\lambda (y + \sqrt{x^2+y^2})}{\lambda x} \)
\( = \lambda^0 \left(\frac{y + \sqrt{x^2+y^2}}{x}\right) \)
\( = \lambda^0 f(x, y) \)
Since \( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \), the function \( f(x, y) \) is a homogeneous function of degree zero.
To solve the differential equation, we make the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into the equation:
\( v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2+(vx)^2}}{x} \)
\( v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2+v^2 x^2}}{x} \)
\( v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2(1+v^2)}}{x} \)
\( v + x \frac{dv}{dx} = \frac{vx + |x|\sqrt{1+v^2}}{x} \)
Assuming \( x > 0 \), then \( |x| = x \):
\( v + x \frac{dv}{dx} = \frac{vx + x\sqrt{1+v^2}}{x} \)
\( v + x \frac{dv}{dx} = v + \sqrt{1+v^2} \)
Now, separate the variables:
\( x \frac{dv}{dx} = \sqrt{1+v^2} \)
\( \frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x} \)
Now, integrate both sides:
\( \int \frac{dv}{\sqrt{1+v^2}} = \int \frac{1}{x} dx \)
Using the standard integral formula \( \int \frac{1}{\sqrt{a^2+x^2}} dx = \log|x + \sqrt{a^2+x^2}| \), where \( a=1 \):
\( \log|v + \sqrt{1+v^2}| = \log|x| + \log C \)
\( \log|v + \sqrt{1+v^2}| = \log|Cx| \)
\( v + \sqrt{1+v^2} = Cx \)
Substitute back \( v = \frac{y}{x} \):
\( \frac{y}{x} + \sqrt{1+\left(\frac{y}{x}\right)^2} = Cx \)
\( \frac{y}{x} + \sqrt{1+\frac{y^2}{x^2}} = Cx \)
\( \frac{y}{x} + \sqrt{\frac{x^2+y^2}{x^2}} = Cx \)
\( \frac{y}{x} + \frac{\sqrt{x^2+y^2}}{|x|} = Cx \)
Assuming \( x > 0 \), then \( |x| = x \):
\( \frac{y + \sqrt{x^2+y^2}}{x} = Cx \)
\( y + \sqrt{x^2+y^2} = Cx^2 \)
This is the general solution.

In simple words: We first put the equation into the \( \frac{dy}{dx} \) format and verified its homogeneous nature. Then, we used the substitution \( y = vx \) to simplify the equation. After separating the \( v \) and \( x \) terms, we integrated both sides. Finally, we substituted \( \frac{y}{x} \) back for \( v \) to get the final solution.

Exam Tip: When dealing with square roots like \( \sqrt{x^2+y^2} \), remember that \( \sqrt{x^2(1+v^2)} = |x|\sqrt{1+v^2} \). Often, it's assumed \( x>0 \) in these problems, allowing you to replace \( |x| \) with \( x \). Be sure to use the correct integration formula for \( \int \frac{dv}{\sqrt{1+v^2}} \).

Question 7. \( \left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\} y dx = \left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\} x dy \)
Answer: We have the given differential equation:
\( \left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\} y dx = \left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\} x dy \)
Rearrange to express in the form \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{\left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\} y}{\left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\} x} = f(x, y) \)
To show that \( f(x, y) \) is a homogeneous function, we replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{\left\{\lambda x \cos\left(\frac{\lambda y}{\lambda x}\right) + \lambda y \sin\left(\frac{\lambda y}{\lambda x}\right)\right\} \lambda y}{\left\{\lambda y \sin\left(\frac{\lambda y}{\lambda x}\right) - \lambda x \cos\left(\frac{\lambda y}{\lambda x}\right)\right\} \lambda x} \)
\( = \frac{\lambda^2 \left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\} y}{\lambda^2 \left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\} x} \)
\( = \lambda^0 \left(\frac{\left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\} y}{\left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\} x}\right) \)
\( = \lambda^0 f(x, y) \)
Since \( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \), the function \( f(x, y) \) is a homogeneous function of degree zero.
To solve the differential equation, we make the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into the equation:
\( v + x \frac{dv}{dx} = \frac{\left\{x \cos\left(\frac{vx}{x}\right) + vx \sin\left(\frac{vx}{x}\right)\right\} vx}{\left\{vx \sin\left(\frac{vx}{x}\right) - x \cos\left(\frac{vx}{x}\right)\right\} x} \)
\( v + x \frac{dv}{dx} = \frac{\left\{x \cos v + vx \sin v\right\} vx}{\left\{vx \sin v - x \cos v\right\} x} \)
\( v + x \frac{dv}{dx} = \frac{x v(\cos v + v \sin v)}{x^2 (v \sin v - \cos v)} \)
\( v + x \frac{dv}{dx} = \frac{v(\cos v + v \sin v)}{v \sin v - \cos v} \)
Now, separate the variables:
\( x \frac{dv}{dx} = \frac{v(\cos v + v \sin v)}{v \sin v - \cos v} - v \)
\( x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v(v \sin v - \cos v)}{v \sin v - \cos v} \)
\( x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v} \)
\( x \frac{dv}{dx} = \frac{2v \cos v}{v \sin v - \cos v} \)
\( \frac{v \sin v - \cos v}{2v \cos v} dv = \frac{dx}{x} \)
\( \left(\frac{v \sin v}{2v \cos v} - \frac{\cos v}{2v \cos v}\right) dv = \frac{dx}{x} \)
\( \left(\frac{1}{2} \tan v - \frac{1}{2v}\right) dv = \frac{dx}{x} \)
Now, integrate both sides:
\( \int \left(\frac{1}{2} \tan v - \frac{1}{2v}\right) dv = \int \frac{1}{x} dx \)
\( \frac{1}{2} \int \tan v dv - \frac{1}{2} \int \frac{1}{v} dv = \int \frac{1}{x} dx \)
\( \frac{1}{2} \log|\sec v| - \frac{1}{2} \log|v| = \log|x| + \log C \)
\( \frac{1}{2} (\log|\sec v| - \log|v|) = \log|x| + \log C \)
\( \frac{1}{2} \log\left|\frac{\sec v}{v}\right| = \log|Cx| \)
\( \log\left|\frac{\sec v}{v}\right|^{1/2} = \log|Cx| \)
\( \sqrt{\frac{\sec v}{v}} = Cx \)
Substitute back \( v = \frac{y}{x} \):
\( \sqrt{\frac{\sec(\frac{y}{x})}{\frac{y}{x}}} = Cx \)
\( \sqrt{\frac{x \sec(\frac{y}{x})}{y}} = Cx \)
Squaring both sides:
\( \frac{x \sec(\frac{y}{x})}{y} = C^2 x^2 \)
\( \sec\left(\frac{y}{x}\right) = C^2 xy \)
\( \frac{1}{\cos(\frac{y}{x})} = C^2 xy \)
\( 1 = C^2 xy \cos\left(\frac{y}{x}\right) \)
Let \( K = C^2 \).
\( K xy \cos\left(\frac{y}{x}\right) = 1 \)
This is the general solution.

In simple words: This was a complex homogeneous equation. We first wrote it in \( \frac{dy}{dx} \) form and proved its homogeneity. Using \( y = vx \) to replace \( y \) and \( \frac{dy}{dx} \), we simplified the equation. After separating the \( v \) and \( x \) terms, we integrated them, using known formulas for tan and log. Finally, we substituted \( \frac{y}{x} \) back for \( v \) and simplified the answer.

Exam Tip: For complex trigonometric homogeneous equations, the algebra after substitution can be lengthy. Carefully simplify terms, use trigonometric identities like \( \sec v = \frac{1}{\cos v} \), and remember to include the constant of integration. Double-check the integration of \( \tan v \) and \( \frac{1}{v} \).

Question 8. \( x \frac{dy}{dx} - y + x \sin \frac{y}{x} = 0 \)
Answer: We have the given differential equation \( x \frac{dy}{dx} - y + x \sin \frac{y}{x} = 0 \).
Rearrange to express in the form \( \frac{dy}{dx} \):
\( x \frac{dy}{dx} = y - x \sin \frac{y}{x} \)
\( \frac{dy}{dx} = \frac{y - x \sin \frac{y}{x}}{x} \)
\( \frac{dy}{dx} = \frac{y}{x} - \sin \frac{y}{x} = f(x, y) \)
To show that \( f(x, y) \) is a homogeneous function, we replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} - \sin \left(\frac{\lambda y}{\lambda x}\right) \)
\( = \frac{y}{x} - \sin \left(\frac{y}{x}\right) \)
\( = \lambda^0 \left(\frac{y}{x} - \sin \frac{y}{x}\right) \)
\( = \lambda^0 f(x, y) \)
Since \( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \), the function \( f(x, y) \) is a homogeneous function of degree zero.
To solve the differential equation, we make the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into the equation:
\( v + x \frac{dv}{dx} = \frac{vx}{x} - \sin \left(\frac{vx}{x}\right) \)
\( v + x \frac{dv}{dx} = v - \sin v \)
Now, separate the variables:
\( x \frac{dv}{dx} = v - \sin v - v \)
\( x \frac{dv}{dx} = -\sin v \)
\( \frac{dv}{\sin v} = -\frac{dx}{x} \)
\( \csc v dv = -\frac{dx}{x} \)
Now, integrate both sides:
\( \int \csc v dv = -\int \frac{1}{x} dx \)
\( \log|\csc v - \cot v| = -\log|x| + \log C \)
\( \log|\csc v - \cot v| = \log\left|\frac{C}{x}\right| \)
\( \csc v - \cot v = \frac{C}{x} \)
Substitute back \( v = \frac{y}{x} \):
\( \csc\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right) = \frac{C}{x} \)
We can also write this using sine and cosine:
\( \frac{1}{\sin(\frac{y}{x})} - \frac{\cos(\frac{y}{x})}{\sin(\frac{y}{x})} = \frac{C}{x} \)
\( \frac{1 - \cos(\frac{y}{x})}{\sin(\frac{y}{x})} = \frac{C}{x} \)
\( x \left(1 - \cos\left(\frac{y}{x}\right)\right) = C \sin\left(\frac{y}{x}\right) \)
This is the general solution.

In simple words: First, we rearranged the equation into the \( \frac{dy}{dx} \) form. We then confirmed it was a homogeneous equation. Using the substitution \( y = vx \), we simplified the equation to separate the \( v \) and \( x \) terms. We integrated both sides using standard integral formulas. Finally, we substituted \( \frac{y}{x} \) back for \( v \) and simplified the expression to get the solution.

Exam Tip: Be familiar with the integral of \( \csc v \), which is \( \log|\csc v - \cot v| \). After substitution, it's often helpful to convert the trigonometric terms back to sine and cosine to simplify the expression further.

Question 9. \( y dx + x \log \left(\frac{y}{x}\right) dy - 2x dy = 0 \)
Answer: We have the given differential equation \( y dx + x \log \left(\frac{y}{x}\right) dy - 2x dy = 0 \).
Rearrange to express in the form \( \frac{dy}{dx} \):
\( y dx = \left(2x - x \log \left(\frac{y}{x}\right)\right) dy \)
\( \frac{dx}{dy} = \frac{2x - x \log \left(\frac{y}{x}\right)}{y} \)
\( \frac{dx}{dy} = \frac{x \left(2 - \log \left(\frac{y}{x}\right)\right)}{y} \)
This equation is of the form \( \frac{dx}{dy} = g\left(\frac{x}{y}\right) \). So, it's a homogeneous equation. We will use the substitution \( x = vy \) instead of \( y = vx \).
Let \( x = vy \), then \( \frac{dx}{dy} = v + y \frac{dv}{dy} \).
Substitute \( x = vy \) and \( \frac{dx}{dy} \) into the equation:
\( v + y \frac{dv}{dy} = \frac{vy \left(2 - \log \left(\frac{vy}{y}\right)\right)}{y} \)
\( v + y \frac{dv}{dy} = v \left(2 - \log v\right) \)
\( v + y \frac{dv}{dy} = 2v - v \log v \)
Now, separate the variables:
\( y \frac{dv}{dy} = 2v - v \log v - v \)
\( y \frac{dv}{dy} = v - v \log v \)
\( y \frac{dv}{dy} = v(1 - \log v) \)
\( \frac{dv}{v(1 - \log v)} = \frac{dy}{y} \)
Now, integrate both sides:
\( \int \frac{dv}{v(1 - \log v)} = \int \frac{1}{y} dy \)
For the left integral, let \( t = 1 - \log v \). Then \( dt = -\frac{1}{v} dv \), so \( \frac{1}{v} dv = -dt \).
\( \int \frac{-dt}{t} = \int \frac{1}{y} dy \)
\( -\log|t| = \log|y| + \log C \)
\( -\log|1 - \log v| = \log|Cy| \)
\( \log|1 - \log v|^{-1} = \log|Cy| \)
\( \frac{1}{|1 - \log v|} = Cy \)
\( \frac{1}{1 - \log v} = \pm Cy \)
Let \( \pm C = K \), a new constant.
\( 1 - \log v = \frac{1}{Ky} \)
\( \log v = 1 - \frac{1}{Ky} \)
Substitute back \( v = \frac{x}{y} \):
\( \log\left(\frac{x}{y}\right) = 1 - \frac{1}{Ky} \)
This is the general solution.

In simple words: This equation was homogeneous, but in terms of \( \frac{dx}{dy} \) and \( \frac{x}{y} \). So, we used the substitution \( x = vy \) and its derivative. After simplifying and separating the \( v \) and \( y \) terms, we integrated both sides. A substitution was needed for the \( v \) integral. Finally, we replaced \( v \) with \( \frac{x}{y} \) to get the solution.

Exam Tip: Be careful to identify whether the differential equation is homogeneous in terms of \( \frac{dy}{dx} \) or \( \frac{dx}{dy} \). This determines whether to use \( y=vx \) or \( x=vy \) substitution. Also, remember the substitution rule for integration, especially for logarithmic terms.

Question 10. \( (1 + e^{\frac{x}{y}})dx + e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)dy = 0 \)
Answer: We have the given differential equation \( (1 + e^{\frac{x}{y}})dx + e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)dy = 0 \).
Rearrange to express in the form \( \frac{dx}{dy} \):
\( (1 + e^{\frac{x}{y}})dx = -e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)dy \)
\( \frac{dx}{dy} = \frac{-e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)}{1 + e^{\frac{x}{y}}} = g\left(\frac{x}{y}\right) \)
This equation is of the form \( \frac{dx}{dy} = g\left(\frac{x}{y}\right) \). So, it's a homogeneous equation. We will use the substitution \( x = vy \) instead of \( y = vx \).
Let \( x = vy \), then \( \frac{dx}{dy} = v + y \frac{dv}{dy} \).
Substitute \( x = vy \) and \( \frac{dx}{dy} \) into the equation:
\( v + y \frac{dv}{dy} = \frac{-e^{\frac{vy}{y}}\left(1 - \frac{vy}{y}\right)}{1 + e^{\frac{vy}{y}}} \)
\( v + y \frac{dv}{dy} = \frac{-e^v(1 - v)}{1 + e^v} \)
Now, separate the variables:
\( y \frac{dv}{dy} = \frac{-e^v(1 - v)}{1 + e^v} - v \)
\( y \frac{dv}{dy} = \frac{-e^v + ve^v - v(1 + e^v)}{1 + e^v} \)
\( y \frac{dv}{dy} = \frac{-e^v + ve^v - v - ve^v}{1 + e^v} \)
\( y \frac{dv}{dy} = \frac{-e^v - v}{1 + e^v} \)
\( y \frac{dv}{dy} = -\frac{v + e^v}{1 + e^v} \)
\( \frac{1 + e^v}{v + e^v} dv = -\frac{dy}{y} \)
Now, integrate both sides:
\( \int \frac{1 + e^v}{v + e^v} dv = -\int \frac{1}{y} dy \)
For the left integral, let \( t = v + e^v \). Then \( dt = (1 + e^v) dv \).
\( \int \frac{dt}{t} = -\int \frac{1}{y} dy \)
\( \log|t| = -\log|y| + \log C \)
\( \log|v + e^v| = \log\left|\frac{C}{y}\right| \)
\( v + e^v = \frac{C}{y} \)
Substitute back \( v = \frac{x}{y} \):
\( \frac{x}{y} + e^{\frac{x}{y}} = \frac{C}{y} \)
Multiply the entire equation by \( y \):
\( x + y e^{\frac{x}{y}} = C \)
This is the general solution.

In simple words: This equation was given in a form that suggested using \( \frac{dx}{dy} \) and \( \frac{x}{y} \). We confirmed its homogeneous nature. We then used the substitution \( x = vy \) to simplify the equation. After separating the \( v \) and \( y \) terms, we integrated both sides. A substitution was helpful for the \( v \) integral. Finally, we replaced \( v \) with \( \frac{x}{y} \) to get the final solution.

Exam Tip: For equations involving \( e^{\frac{x}{y}} \) or \( e^{\frac{y}{x}} \), the appropriate substitution is usually \( x=vy \) or \( y=vx \) respectively. Recognizing the integral of the form \( \int \frac{f'(x)}{f(x)} dx \) is crucial for efficiency in solving these types of problems.

Question 11. (x + y)dy + (x - y)dx = 0, y = 1, when x = 1.
Answer: We are given the differential equation \( (x + y)dy + (x - y)dx = 0 \).
We can rearrange this equation to find \( \frac{dy}{dx} \):
\( (x + y)dy = -(x - y)dx \)
\( \implies \frac{dy}{dx} = -\frac{x - y}{x + y} = \frac{y - x}{y + x} \).
Let \( f(x, y) = \frac{y - x}{y + x} \).
Replacing \( x \) by \( \lambda x \) and \( y \) by \( \lambda y \), we get:
\( f(\lambda x, \lambda y) = \frac{\lambda y - \lambda x}{\lambda y + \lambda x} = \frac{\lambda (y - x)}{\lambda (y + x)} = \frac{y - x}{y + x} = \lambda^0 f(x, y) \).
Since \( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \), the function \( f(x, y) \) is a homogeneous function of degree zero.
To solve this homogeneous equation, we substitute \( y = vx \), which gives \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substituting these into the differential equation:
\( v + x \frac{dv}{dx} = \frac{vx - x}{vx + x} \)
\( \implies v + x \frac{dv}{dx} = \frac{x(v - 1)}{x(v + 1)} \)
\( \implies v + x \frac{dv}{dx} = \frac{v - 1}{v + 1} \)
\( \implies x \frac{dv}{dx} = \frac{v - 1}{v + 1} - v \)
\( \implies x \frac{dv}{dx} = \frac{v - 1 - v(v + 1)}{v + 1} \)
\( \implies x \frac{dv}{dx} = \frac{v - 1 - v^2 - v}{v + 1} \)
\( \implies x \frac{dv}{dx} = \frac{-1 - v^2}{v + 1} \)
\( \implies x \frac{dv}{dx} = -\frac{v^2 + 1}{v + 1} \).
Now, we separate the variables:
\( \frac{v + 1}{v^2 + 1} dv = -\frac{dx}{x} \).
Integrate both sides:
\( \int \frac{v + 1}{v^2 + 1} dv = -\int \frac{dx}{x} \)
\( \implies \int \frac{v}{v^2 + 1} dv + \int \frac{1}{v^2 + 1} dv = -\int \frac{dx}{x} \).
For the first integral, let \( t = v^2 + 1 \), so \( dt = 2v dv \implies v dv = \frac{1}{2} dt \).
\( \implies \frac{1}{2} \int \frac{dt}{t} + \int \frac{1}{v^2 + 1} dv = -\int \frac{dx}{x} \)
\( \implies \frac{1}{2} \log|t| + \arctan(v) = -\log|x| + C \)
Substitute back \( t = v^2 + 1 \):
\( \implies \frac{1}{2} \log(v^2 + 1) + \arctan(v) = -\log|x| + C \).
Now substitute \( v = \frac{y}{x} \):
\( \implies \frac{1}{2} \log\left(\left(\frac{y}{x}\right)^2 + 1\right) + \arctan\left(\frac{y}{x}\right) = -\log|x| + C \)
\( \implies \frac{1}{2} \log\left(\frac{y^2 + x^2}{x^2}\right) + \arctan\left(\frac{y}{x}\right) = -\log|x| + C \)
\( \implies \frac{1}{2} (\log(y^2 + x^2) - \log(x^2)) + \arctan\left(\frac{y}{x}\right) = -\log|x| + C \)
\( \implies \frac{1}{2} \log(x^2 + y^2) - \log|x| + \arctan\left(\frac{y}{x}\right) = -\log|x| + C \)
\( \implies \frac{1}{2} \log(x^2 + y^2) + \arctan\left(\frac{y}{x}\right) = C \).
We are given the condition \( y = 1 \) when \( x = 1 \). Substitute these values to find \( C \):
\( \frac{1}{2} \log(1^2 + 1^2) + \arctan\left(\frac{1}{1}\right) = C \)
\( \implies \frac{1}{2} \log(2) + \arctan(1) = C \)
\( \implies \frac{1}{2} \log(2) + \frac{\pi}{4} = C \).
Therefore, the particular solution is:
\( \frac{1}{2} \log(x^2 + y^2) + \arctan\left(\frac{y}{x}\right) = \frac{1}{2} \log(2) + \frac{\pi}{4} \).
In simple words: First, we change the given equation into a standard form. Then, we check if it is homogeneous, which means all terms have the same total power. We use a special trick by replacing \( y \) with \( vx \) to simplify the equation, then separate the \( v \) and \( x \) terms. After integrating both sides, we put \( y/x \) back in for \( v \). Finally, we use the given starting values for \( x \) and \( y \) to find the constant \( C \), giving us the exact solution for those specific conditions.

Exam Tip: Remember to always check for homogeneity first, then use the substitution \( y = vx \) for \( dy/dx \) type equations or \( x = vy \) for \( dx/dy \) type equations. Don't forget to evaluate the constant of integration \( C \) using the given initial conditions to find the particular solution.

Question 12. x²dy + (xy + y²)dx = 0, y = 1, when x = 1.
Answer: We are given the differential equation \( x^2 dy + (xy + y^2) dx = 0 \).
First, we rearrange it to find \( \frac{dy}{dx} \):
\( x^2 dy = -(xy + y^2) dx \)
\( \implies \frac{dy}{dx} = -\frac{xy + y^2}{x^2} \).
Let \( f(x, y) = -\frac{xy + y^2}{x^2} \).
To check if it is homogeneous, we replace \( x \) by \( \lambda x \) and \( y \) by \( \lambda y \):
\( f(\lambda x, \lambda y) = -\frac{(\lambda x)(\lambda y) + (\lambda y)^2}{(\lambda x)^2} \)
\( \implies f(\lambda x, \lambda y) = -\frac{\lambda^2 xy + \lambda^2 y^2}{\lambda^2 x^2} \)
\( \implies f(\lambda x, \lambda y) = -\frac{\lambda^2 (xy + y^2)}{\lambda^2 x^2} \)
\( \implies f(\lambda x, \lambda y) = -\frac{xy + y^2}{x^2} = \lambda^0 f(x, y) \).
Since \( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \), the differential equation is homogeneous.
Now, we use the substitution \( y = vx \), which implies \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substituting these into the equation:
\( v + x \frac{dv}{dx} = -\frac{x(vx) + (vx)^2}{x^2} \)
\( \implies v + x \frac{dv}{dx} = -\frac{vx^2 + v^2 x^2}{x^2} \)
\( \implies v + x \frac{dv}{dx} = -\frac{x^2(v + v^2)}{x^2} \)
\( \implies v + x \frac{dv}{dx} = -(v + v^2) \)
\( \implies x \frac{dv}{dx} = -v - v^2 - v \)
\( \implies x \frac{dv}{dx} = -(2v + v^2) \).
Separating the variables, we get:
\( \frac{dv}{v^2 + 2v} = -\frac{dx}{x} \).
Integrate both sides:
\( \int \frac{dv}{v(v + 2)} = -\int \frac{dx}{x} \).
We use partial fraction decomposition for the left side:
\( \frac{1}{v(v + 2)} = \frac{A}{v} + \frac{B}{v + 2} \)
\( \implies 1 = A(v + 2) + Bv \).
If \( v = 0 \), then \( 1 = 2A \implies A = \frac{1}{2} \).
If \( v = -2 \), then \( 1 = -2B \implies B = -\frac{1}{2} \).
So, \( \int \left(\frac{1/2}{v} - \frac{1/2}{v + 2}\right) dv = -\int \frac{dx}{x} \)
\( \implies \frac{1}{2} \int \left(\frac{1}{v} - \frac{1}{v + 2}\right) dv = -\int \frac{dx}{x} \)
\( \implies \frac{1}{2} (\log|v| - \log|v + 2|) = -\log|x| + C' \)
\( \implies \frac{1}{2} \log\left|\frac{v}{v + 2}\right| = -\log|x| + C' \)
\( \implies \log\left|\frac{v}{v + 2}\right| = -2\log|x| + 2C' \)
\( \implies \log\left|\frac{v}{v + 2}\right| = \log|x^{-2}| + \log|K| \) (where \( \log K = 2C' \))
\( \implies \frac{v}{v + 2} = \frac{K}{x^2} \).
Substitute back \( v = \frac{y}{x} \):
\( \frac{y/x}{y/x + 2} = \frac{K}{x^2} \)
\( \implies \frac{y/x}{(y + 2x)/x} = \frac{K}{x^2} \)
\( \implies \frac{y}{y + 2x} = \frac{K}{x^2} \)
\( \implies x^2 y = K(y + 2x) \).
We are given \( y = 1 \) when \( x = 1 \). Substitute these values to find \( K \):
\( (1)^2 (1) = K(1 + 2(1)) \)
\( \implies 1 = K(3) \)
\( \implies K = \frac{1}{3} \).
So, the particular solution is:
\( x^2 y = \frac{1}{3}(y + 2x) \)
\( \implies 3x^2 y = y + 2x \).
In simple words: We first rewrite the equation to find \( dy/dx \). Then, we check if it's homogeneous by changing \( x \) and \( y \) with \( \lambda x \) and \( \lambda y \). Since it is, we replace \( y \) with \( vx \) and \( dy/dx \) with \( v + x(dv/dx) \). This lets us separate the \( v \) and \( x \) parts, which we can then integrate. After integrating, we change \( v \) back to \( y/x \) and use the given initial values for \( x \) and \( y \) to calculate the constant \( K \), giving us the final specific solution.

Exam Tip: For problems involving integration of rational functions like \( \frac{1}{v^2 + 2v} \), remember to use partial fraction decomposition to simplify the integrand before integrating. This is a common step in solving homogeneous differential equations.

Question 13. (x sin² \( \frac{y}{x} \) – y) dx + x dy = 0, y = \( \frac{\pi}{4} \), when x = 1.
Answer: We are given the differential equation \( (x \sin^2 \left(\frac{y}{x}\right) - y) dx + x dy = 0 \).
Rearranging to find \( \frac{dy}{dx} \):
\( x dy = -(x \sin^2 \left(\frac{y}{x}\right) - y) dx \)
\( \implies \frac{dy}{dx} = -\frac{x \sin^2 \left(\frac{y}{x}\right) - y}{x} \)
\( \implies \frac{dy}{dx} = \frac{y - x \sin^2 \left(\frac{y}{x}\right)}{x} \).
Let \( f(x, y) = \frac{y - x \sin^2 \left(\frac{y}{x}\right)}{x} \).
To check for homogeneity, replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{\lambda y - \lambda x \sin^2 \left(\frac{\lambda y}{\lambda x}\right)}{\lambda x} \)
\( \implies f(\lambda x, \lambda y) = \frac{\lambda y - \lambda x \sin^2 \left(\frac{y}{x}\right)}{\lambda x} \)
\( \implies f(\lambda x, \lambda y) = \frac{\lambda \left(y - x \sin^2 \left(\frac{y}{x}\right)\right)}{\lambda x} \)
\( \implies f(\lambda x, \lambda y) = \frac{y - x \sin^2 \left(\frac{y}{x}\right)}{x} = \lambda^0 f(x, y) \).
The equation is homogeneous.
Substitute \( y = vx \), so \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
\( v + x \frac{dv}{dx} = \frac{vx - x \sin^2(v)}{x} \)
\( \implies v + x \frac{dv}{dx} = v - \sin^2(v) \)
\( \implies x \frac{dv}{dx} = -\sin^2(v) \).
Separating variables:
\( \frac{dv}{\sin^2(v)} = -\frac{dx}{x} \)
\( \implies \csc^2(v) dv = -\frac{dx}{x} \).
Integrate both sides:
\( \int \csc^2(v) dv = -\int \frac{dx}{x} \)
\( \implies -\cot(v) = -\log|x| + C \)
\( \implies \cot(v) = \log|x| - C \).
Substitute back \( v = \frac{y}{x} \):
\( \cot\left(\frac{y}{x}\right) = \log|x| - C \).
We are given \( y = \frac{\pi}{4} \) when \( x = 1 \). Substitute these values:
\( \cot\left(\frac{\pi/4}{1}\right) = \log|1| - C \)
\( \implies \cot\left(\frac{\pi}{4}\right) = 0 - C \)
\( \implies 1 = -C \implies C = -1 \).
So, the particular solution is:
\( \cot\left(\frac{y}{x}\right) = \log|x| - (-1) \)
\( \implies \cot\left(\frac{y}{x}\right) = \log|x| + 1 \).
In simple words: We begin by changing the equation into the \( dy/dx \) form. We then check if it's homogeneous, which means it stays the same when we multiply \( x \) and \( y \) by a constant. Because it is, we replace \( y \) with \( vx \) and \( dy/dx \) with \( v + x(dv/dx) \) to make it simpler. After this, we separate the \( v \) and \( x \) terms and integrate them. Finally, we put \( y/x \) back for \( v \) and use the given starting values for \( x \) and \( y \) to find the constant, giving us the exact answer for those specific conditions.

Exam Tip: When integrating trigonometric functions, ensure you remember standard integrals like \( \int \csc^2(v) dv = -\cot(v) \). Also, accurately handling the constant of integration and applying initial conditions are crucial for finding the correct particular solution.

Question 14. \( \frac{dy}{dx} – \frac{y}{x} + \csc \left(\frac{y}{x}\right) = 0 \), y = 0, when x = 1.
Answer: We are given the differential equation \( \frac{dy}{dx} – \frac{y}{x} + \csc \left(\frac{y}{x}\right) = 0 \).
Rearranging it:
\( \frac{dy}{dx} = \frac{y}{x} - \csc \left(\frac{y}{x}\right) \).
Let \( f(x, y) = \frac{y}{x} - \csc \left(\frac{y}{x}\right) \).
To check for homogeneity, replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} - \csc \left(\frac{\lambda y}{\lambda x}\right) \)
\( \implies f(\lambda x, \lambda y) = \frac{y}{x} - \csc \left(\frac{y}{x}\right) = \lambda^0 f(x, y) \).
The equation is homogeneous.
Substitute \( y = vx \), so \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
\( v + x \frac{dv}{dx} = v - \csc(v) \)
\( \implies x \frac{dv}{dx} = -\csc(v) \).
Separating variables:
\( \frac{dv}{\csc(v)} = -\frac{dx}{x} \)
\( \implies \sin(v) dv = -\frac{dx}{x} \).
Integrate both sides:
\( \int \sin(v) dv = -\int \frac{dx}{x} \)
\( \implies -\cos(v) = -\log|x| + C' \)
\( \implies \cos(v) = \log|x| - C' \). Let \( C = -C' \).
\( \implies \cos(v) = \log|x| + C \).
Substitute back \( v = \frac{y}{x} \):
\( \cos\left(\frac{y}{x}\right) = \log|x| + C \).
We are given \( y = 0 \) when \( x = 1 \). Substitute these values:
\( \cos\left(\frac{0}{1}\right) = \log|1| + C \)
\( \implies \cos(0) = 0 + C \)
\( \implies 1 = C \).
So, the particular solution is:
\( \cos\left(\frac{y}{x}\right) = \log|x| + 1 \).
This can also be written as \( \log|x| = \cos\left(\frac{y}{x}\right) - 1 \).
In simple words: We first rewrite the equation to clearly see its form. Since it's homogeneous, meaning all terms have the same overall power, we can use the substitution \( y = vx \). This changes the equation into one where we can easily separate the variables \( v \) and \( x \). After integrating both parts, we replace \( v \) with \( y/x \). Finally, we use the given starting conditions for \( x \) and \( y \) to find the constant value \( C \), which gives us the specific solution for this problem.

Exam Tip: Be careful with signs when integrating and rearranging terms. A common mistake is misplacing a negative sign, which can lead to an incorrect solution. Double-check your algebraic manipulations and integration results.

Question 15. 2 xy + y² - 2x² \( \frac{dy}{dx} \) = 0, y = 2, when x = 1.
Answer: We are given the differential equation \( 2xy + y^2 - 2x^2 \frac{dy}{dx} = 0 \).
Rearranging to find \( \frac{dy}{dx} \):
\( 2x^2 \frac{dy}{dx} = 2xy + y^2 \)
\( \implies \frac{dy}{dx} = \frac{2xy + y^2}{2x^2} \).
Let \( f(x, y) = \frac{2xy + y^2}{2x^2} \).
To check for homogeneity, replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{2(\lambda x)(\lambda y) + (\lambda y)^2}{2(\lambda x)^2} \)
\( \implies f(\lambda x, \lambda y) = \frac{2\lambda^2 xy + \lambda^2 y^2}{2\lambda^2 x^2} \)
\( \implies f(\lambda x, \lambda y) = \frac{\lambda^2 (2xy + y^2)}{\lambda^2 (2x^2)} \)
\( \implies f(\lambda x, \lambda y) = \frac{2xy + y^2}{2x^2} = \lambda^0 f(x, y) \).
The equation is homogeneous.
Substitute \( y = vx \), so \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
\( v + x \frac{dv}{dx} = \frac{2x(vx) + (vx)^2}{2x^2} \)
\( \implies v + x \frac{dv}{dx} = \frac{2vx^2 + v^2 x^2}{2x^2} \)
\( \implies v + x \frac{dv}{dx} = \frac{x^2(2v + v^2)}{2x^2} \)
\( \implies v + x \frac{dv}{dx} = \frac{2v + v^2}{2} \)
\( \implies x \frac{dv}{dx} = \frac{2v + v^2}{2} - v \)
\( \implies x \frac{dv}{dx} = \frac{2v + v^2 - 2v}{2} \)
\( \implies x \frac{dv}{dx} = \frac{v^2}{2} \).
Separating variables:
\( \frac{2 dv}{v^2} = \frac{dx}{x} \)
\( \implies 2 v^{-2} dv = \frac{dx}{x} \).
Integrate both sides:
\( \int 2 v^{-2} dv = \int \frac{dx}{x} \)
\( \implies 2 \frac{v^{-1}}{-1} = \log|x| + C \)
\( \implies -\frac{2}{v} = \log|x| + C \).
Substitute back \( v = \frac{y}{x} \):
\( -\frac{2}{y/x} = \log|x| + C \)
\( \implies -\frac{2x}{y} = \log|x| + C \).
We are given \( y = 2 \) when \( x = 1 \). Substitute these values:
\( -\frac{2(1)}{2} = \log|1| + C \)
\( \implies -1 = 0 + C \)
\( \implies C = -1 \).
So, the particular solution is:
\( -\frac{2x}{y} = \log|x| - 1 \)
\( \implies 1 - \log|x| = \frac{2x}{y} \)
\( \implies y = \frac{2x}{1 - \log|x|} \).
In simple words: We first rewrite the equation to find \( dy/dx \). After checking that it is homogeneous, we use the substitution \( y = vx \). This allows us to separate the variables \( v \) and \( x \), making the equation easier to integrate. Once integrated, we substitute \( v = y/x \) back into the solution. Finally, we use the given starting values for \( x \) and \( y \) to determine the constant of integration, giving us the specific solution for the problem.

Exam Tip: Pay close attention to the algebraic manipulation when separating variables. Incorrectly grouping terms or signs can lead to errors in integration and, ultimately, the final solution. Remember that \( \int v^{-2} dv = -v^{-1} \).

Question 16. A homogeneous differential equation of the form \( \frac{dx}{dy} = h\left(\frac{x}{y}\right) \) can be solved by making the substitution.
(a) y = vx
(b) v = yx
(c) x = vy
(d) x = v
Answer: (c) x = vy
In simple words: When a homogeneous differential equation looks like \( dx/dy = h(x/y) \), it means the right side only depends on the ratio \( x/y \). In this situation, the correct way to simplify and solve it is by replacing \( x \) with \( vy \).

Exam Tip: Remember the rule: if the equation is in the form \( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \), substitute \( y = vx \). If it's in the form \( \frac{dx}{dy} = f\left(\frac{x}{y}\right) \), substitute \( x = vy \). This is a fundamental concept for solving homogeneous differential equations.

Question 17. Which of the following is a homogeneous differential equation?
(a) (4x + 6y + 5)dy – (3y + 2x + 4) dx = 0
(b) xy dx – (x³ + y³)dy = 0
(c) (x³ + 2y²) dx + 2xy dy = 0
(d) y²dx + (x² – xy – y²) dy = 0
Answer: (d) y²dx + (x² – xy – y²) dy = 0
In simple words: A homogeneous differential equation is one where all terms have the same total power. For option (D), if you look at each part, like \( y^2 \) or \( x^2 \), they all have a power of 2. This makes the equation balanced and homogeneous. The other options have terms with different total powers, so they are not homogeneous.

Exam Tip: To identify a homogeneous differential equation, check the sum of powers of \( x \) and \( y \) in each term. If the sum is the same for all terms, the equation is homogeneous. Constant terms (like 5 or 4 in option A) automatically make an equation non-homogeneous.