GSEB Class 12 Maths Solutions Chapter 9 વિકલ સમીકરણો Exercise 9.5

Questions 1 To 10: Show That The Given Differential Equations Are Homogeneous Differential Equations And Find The Solution For Each.

Question 1. \( (x^{2} + xy) dy = (x^{2} + y^{2}) dx \)
Answer: The given differential equation is:
\( (x^{2} + xy) dy = (x^{2} + y^{2}) dx \)
We can rearrange this equation to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{x^{2} + y^{2}}{x^{2} + xy} \) ... (1)
Let's define \( F(x, y) = \frac{x^{2} + y^{2}}{x^{2} + xy} \).
Now, we will replace x with \( \lambda x \) and y with \( \lambda y \) in \( F(x, y) \):
\( F(\lambda x, \lambda y) = \frac{(\lambda x)^{2} + (\lambda y)^{2}}{(\lambda x)^{2} + (\lambda x)(\lambda y)} \)
\( F(\lambda x, \lambda y) = \frac{\lambda^{2}x^{2} + \lambda^{2}y^{2}}{\lambda^{2}x^{2} + \lambda^{2}xy} \)
\( F(\lambda x, \lambda y) = \frac{\lambda^{2}(x^{2} + y^{2})}{\lambda^{2}(x^{2} + xy)} \)
\( F(\lambda x, \lambda y) = \lambda^{0} \frac{x^{2} + y^{2}}{x^{2} + xy} \)
\( F(\lambda x, \lambda y) = \lambda^{0} F(x, y) \)
This shows that \( F(x, y) \) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve this, let's substitute \( y = vx \).
Differentiating \( y = vx \) with respect to x gives:
\( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Now, we will substitute this value of \( \frac{dy}{dx} \) and \( y = vx \) 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, separate the variables:
\( x \frac{dv}{dx} = \frac{1 + v^{2}}{1 + v} - v \)
\( x \frac{dv}{dx} = \frac{1 + v^{2} - v(1 + v)}{1 + v} \)
\( x \frac{dv}{dx} = \frac{1 + v^{2} - v - v^{2}}{1 + v} \)
\( x \frac{dv}{dx} = \frac{1 - v}{1 + v} \)
Now, we will separate the variables and integrate both sides:
\( \frac{1 + v}{1 - v} dv = \frac{dx}{x} \)
To integrate \( \frac{1 + v}{1 - v} \), we can rewrite it as \( \frac{-(v + 1)}{v - 1} \).
So, \( \frac{-(v + 1)}{v - 1} dv = \frac{dx}{x} \)
\( \frac{-(v - 1 + 2)}{v - 1} dv = \frac{dx}{x} \)
\( - \left( \frac{v - 1}{v - 1} + \frac{2}{v - 1} \right) dv = \frac{dx}{x} \)
\( - \left( 1 + \frac{2}{v - 1} \right) dv = \frac{dx}{x} \)
Integrate both sides:
\( - \int \left( 1 + \frac{2}{v - 1} \right) dv = \int \frac{dx}{x} \)
\( - (v + 2 \log|v - 1|) = \log|x| + C_{1} \)
\( - v - 2 \log|v - 1| = \log|x| + C_{1} \)
Substitute back \( v = \frac{y}{x} \):
\( - \frac{y}{x} - 2 \log\left|\frac{y}{x} - 1\right| = \log|x| + C_{1} \)
\( - \frac{y}{x} - 2 \log\left|\frac{y - x}{x}\right| = \log|x| + C_{1} \)
\( - \frac{y}{x} - 2 (\log|y - x| - \log|x|) = \log|x| + C_{1} \)
\( - \frac{y}{x} - 2 \log|y - x| + 2 \log|x| = \log|x| + C_{1} \)
\( - \frac{y}{x} - 2 \log|y - x| + \log|x| = C_{1} \)
\( - \frac{y}{x} + \log|x| = C_{1} + 2 \log|y - x| \)
We can write \( C_{1} \) as \( -\log C \).
\( - \frac{y}{x} + \log|x| = -\log C + 2 \log|y - x| \)
\( \log|x| + \log C = \frac{y}{x} + 2 \log|y - x| \)
\( \log(Cx) = \frac{y}{x} + \log(y - x)^{2} \)
\( \log(Cx) - \log(y - x)^{2} = \frac{y}{x} \)
\( \log \left( \frac{Cx}{(y - x)^{2}} \right) = \frac{y}{x} \)
\( \frac{Cx}{(y - x)^{2}} = e^{\frac{y}{x}} \)
\( Cx = (y - x)^{2} e^{\frac{y}{x}} \)
Therefore, the solution to the differential equation is \( Cx = (y - x)^{2} e^{\frac{y}{x}} \).
In simple words: We began by rewriting the equation to get \( \frac{dy}{dx} \). Then we checked if it was a "homogeneous" type of equation, which it was. To solve it, we made a substitution, changed variables, and then integrated both sides. Finally, we put the original variables back to get the solution.

Exam Tip: Remember to always check for homogeneity first, and for substitution \( y=vx \), derive \( \frac{dy}{dx} \) correctly. Variable separation and proper integration are key steps.

Question 2. \( y' = \frac{x+y}{x} \)
Answer: The given differential equation is:
\( y' = \frac{x+y}{x} \)
We can write \( y' \) as \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{x+y}{x} \) ... (1)
Let's define \( F(x, y) = \frac{x+y}{x} \).
Now, we will replace x with \( \lambda x \) and y with \( \lambda y \) in \( F(x, y) \):
\( F(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x} \)
\( F(\lambda x, \lambda y) = \frac{\lambda (x + y)}{\lambda x} \)
\( F(\lambda x, \lambda y) = \lambda^{0} \frac{x+y}{x} \)
\( F(\lambda x, \lambda y) = \lambda^{0} F(x, y) \)
This shows that \( F(x, y) \) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve this, let's substitute \( y = vx \).
Differentiating \( y = vx \) with respect to x gives:
\( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Now, we will substitute this value of \( \frac{dy}{dx} \) and \( y = vx \) into equation (1):
\( v + x \frac{dv}{dx} = \frac{x + vx}{x} \)
\( v + x \frac{dv}{dx} = \frac{x(1 + v)}{x} \)
\( v + x \frac{dv}{dx} = 1 + v \)
Now, separate the variables:
\( x \frac{dv}{dx} = 1 + v - v \)
\( x \frac{dv}{dx} = 1 \)
Now, we will separate the variables and integrate both sides:
\( dv = \frac{dx}{x} \)
\( \int dv = \int \frac{dx}{x} \)
\( v = \log|x| + C \)
Substitute back \( v = \frac{y}{x} \):
\( \frac{y}{x} = \log|x| + C \)
\( y = x \log|x| + Cx \)
Therefore, the solution to the differential equation is \( y = x \log|x| + Cx \).
In simple words: We first rewrote the given equation. Then, we verified if it was a homogeneous equation. By replacing \( y \) with \( vx \) and differentiating, we simplified the equation. After separating the variables, we integrated to find the solution and then replaced \( v \) back with \( \frac{y}{x} \).

Exam Tip: For simple homogeneous equations like this, the algebra after substitution \( y=vx \) often simplifies quickly, making variable separation straightforward. Be careful with the integration constant.

Question 3. \( (x - y) dy - (x + y) dx = 0 \)
Answer: The given differential equation is:
\( (x - y) dy - (x + y) dx = 0 \)
We can rearrange this equation to find \( \frac{dy}{dx} \):
\( (x - y) dy = (x + y) dx \)
\( \frac{dy}{dx} = \frac{x+y}{x-y} \) ... (1)
Let's define \( F(x, y) = \frac{x+y}{x-y} \).
Now, we will replace x with \( \lambda x \) and y with \( \lambda y \) in \( F(x, y) \):
\( F(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x - \lambda y} \)
\( F(\lambda x, \lambda y) = \frac{\lambda (x + y)}{\lambda (x - y)} \)
\( F(\lambda x, \lambda y) = \lambda^{0} \frac{x+y}{x-y} \)
\( F(\lambda x, \lambda y) = \lambda^{0} F(x, y) \)
This shows that \( F(x, y) \) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve this, let's substitute \( y = vx \).
Differentiating \( y = vx \) with respect to x gives:
\( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Now, we will substitute this value of \( \frac{dy}{dx} \) and \( y = vx \) into equation (1):
\( v + x \frac{dv}{dx} = \frac{x + vx}{x - vx} \)
\( v + x \frac{dv}{dx} = \frac{x(1 + v)}{x(1 - v)} \)
\( v + x \frac{dv}{dx} = \frac{1 + v}{1 - v} \)
Now, separate the variables:
\( x \frac{dv}{dx} = \frac{1 + v}{1 - v} - v \)
\( x \frac{dv}{dx} = \frac{1 + v - v(1 - v)}{1 - v} \)
\( x \frac{dv}{dx} = \frac{1 + v - v + v^{2}}{1 - v} \)
\( x \frac{dv}{dx} = \frac{1 + v^{2}}{1 - v} \)
Now, we will separate the variables and integrate both sides:
\( \frac{1 - v}{1 + v^{2}} dv = \frac{dx}{x} \)
\( \int \frac{1 - v}{1 + v^{2}} dv = \int \frac{dx}{x} \)
We can split the integral on the left side:
\( \int \frac{1}{1 + v^{2}} dv - \int \frac{v}{1 + v^{2}} dv = \int \frac{dx}{x} \)
For the second integral, let \( u = 1 + v^{2} \), so \( du = 2v dv \), which means \( v dv = \frac{1}{2} du \).
\( \tan^{-1}(v) - \frac{1}{2} \int \frac{1}{u} du = \log|x| + C \)
\( \tan^{-1}(v) - \frac{1}{2} \log|1 + v^{2}| = \log|x| + C \)
Substitute back \( v = \frac{y}{x} \):
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log\left|1 + \left(\frac{y}{x}\right)^{2}\right| = \log|x| + C \)
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log\left|\frac{x^{2} + y^{2}}{x^{2}}\right| = \log|x| + C \)
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} (\log|x^{2} + y^{2}| - \log|x^{2}|) = \log|x| + C \)
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log(x^{2} + y^{2}) + \frac{1}{2} \log(x^{2}) = \log|x| + C \)
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log(x^{2} + y^{2}) + \log|x| = \log|x| + C \)
\( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log(x^{2} + y^{2}) = C \)
Therefore, the solution to the differential equation is \( \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \log(x^{2} + y^{2}) = C \).
In simple words: We began by rearranging the equation to find \( \frac{dy}{dx} \). We then checked if it was a homogeneous equation. By substituting \( y = vx \), we transformed the equation, separated the variables, and integrated. Remember to split the integral for easier calculation and then put back the original variables.

Exam Tip: When integrating fractions like \( \frac{1-v}{1+v^2} \), split them into separate terms for \( \tan^{-1} \) and \( \log \) forms. Pay close attention to the constant of integration and logarithmic properties.

Question 4. \( (x^{2} - y^{2}) dx + 2xy dy = 0 \)
Answer: The given differential equation is:
\( (x^{2} - y^{2}) dx + 2xy dy = 0 \)
We can rearrange this equation to find \( \frac{dy}{dx} \):
\( 2xy dy = -(x^{2} - y^{2}) dx \)
\( 2xy dy = (y^{2} - x^{2}) dx \)
\( \frac{dy}{dx} = \frac{y^{2} - x^{2}}{2xy} \) ... (1)
Let's define \( F(x, y) = \frac{y^{2} - x^{2}}{2xy} \).
Now, we will replace x with \( \lambda x \) and y with \( \lambda y \) in \( F(x, y) \):
\( F(\lambda x, \lambda y) = \frac{(\lambda y)^{2} - (\lambda x)^{2}}{2(\lambda x)(\lambda y)} \)
\( F(\lambda x, \lambda y) = \frac{\lambda^{2}y^{2} - \lambda^{2}x^{2}}{2\lambda^{2}xy} \)
\( F(\lambda x, \lambda y) = \frac{\lambda^{2}(y^{2} - x^{2})}{\lambda^{2}(2xy)} \)
\( F(\lambda x, \lambda y) = \lambda^{0} \frac{y^{2} - x^{2}}{2xy} \)
\( F(\lambda x, \lambda y) = \lambda^{0} F(x, y) \)
This shows that \( F(x, y) \) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve this, let's substitute \( y = vx \).
Differentiating \( y = vx \) with respect to x gives:
\( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Now, we will substitute this value of \( \frac{dy}{dx} \) and \( y = vx \) into equation (1):
\( v + x \frac{dv}{dx} = \frac{(vx)^{2} - x^{2}}{2x(vx)} \)
\( v + x \frac{dv}{dx} = \frac{v^{2}x^{2} - x^{2}}{2vx^{2}} \)
\( v + x \frac{dv}{dx} = \frac{x^{2}(v^{2} - 1)}{x^{2}(2v)} \)
\( v + x \frac{dv}{dx} = \frac{v^{2} - 1}{2v} \)
Now, separate the variables:
\( x \frac{dv}{dx} = \frac{v^{2} - 1}{2v} - v \)
\( x \frac{dv}{dx} = \frac{v^{2} - 1 - 2v^{2}}{2v} \)
\( x \frac{dv}{dx} = \frac{-v^{2} - 1}{2v} \)
\( x \frac{dv}{dx} = -\frac{v^{2} + 1}{2v} \)
Now, we will separate the variables and integrate both sides:
\( \frac{2v}{v^{2} + 1} dv = -\frac{dx}{x} \)
\( \int \frac{2v}{v^{2} + 1} dv = -\int \frac{dx}{x} \)
For the left side, let \( u = v^{2} + 1 \), so \( du = 2v dv \).
\( \int \frac{1}{u} du = -\log|x| + \log C \)
\( \log|v^{2} + 1| = \log\left|\frac{C}{x}\right| \)
\( v^{2} + 1 = \frac{C}{x} \)
Substitute back \( v = \frac{y}{x} \):
\( \left(\frac{y}{x}\right)^{2} + 1 = \frac{C}{x} \)
\( \frac{y^{2}}{x^{2}} + 1 = \frac{C}{x} \)
\( \frac{y^{2} + x^{2}}{x^{2}} = \frac{C}{x} \)
\( y^{2} + x^{2} = Cx \)
Therefore, the solution to the differential equation is \( x^{2} + y^{2} = Cx \).
In simple words: First, we wrote the equation as \( \frac{dy}{dx} \). We then proved it was a homogeneous equation. By using the substitution \( y = vx \), we converted the equation, separated variables, and integrated. Finally, we substituted \( v \) back to find the solution.

Exam Tip: When \( \frac{f'(v)}{f(v)} \) appears in integration, it directly integrates to \( \log|f(v)| \). Look for opportunities to use this pattern.

Question 5. \( x^{2} \frac{dy}{dx} = x^{2} - 2y^{2} + xy \)
Answer: The given differential equation is:
\( x^{2} \frac{dy}{dx} = x^{2} - 2y^{2} + xy \)
We can rearrange this equation to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{x^{2} - 2y^{2} + xy}{x^{2}} \) ... (1)
Let's define \( F(x, y) = \frac{x^{2} - 2y^{2} + xy}{x^{2}} \).
Now, we will replace x with \( \lambda x \) and y with \( \lambda y \) in \( F(x, y) \):
\( F(\lambda x, \lambda y) = \frac{(\lambda x)^{2} - 2(\lambda y)^{2} + (\lambda x)(\lambda y)}{(\lambda x)^{2}} \)
\( F(\lambda x, \lambda y) = \frac{\lambda^{2}x^{2} - 2\lambda^{2}y^{2} + \lambda^{2}xy}{\lambda^{2}x^{2}} \)
\( F(\lambda x, \lambda y) = \frac{\lambda^{2}(x^{2} - 2y^{2} + xy)}{\lambda^{2}x^{2}} \)
\( F(\lambda x, \lambda y) = \lambda^{0} \frac{x^{2} - 2y^{2} + xy}{x^{2}} \)
\( F(\lambda x, \lambda y) = \lambda^{0} F(x, y) \)
This shows that \( F(x, y) \) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve this, let's substitute \( y = vx \).
Differentiating \( y = vx \) with respect to x gives:
\( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Now, we will substitute this value of \( \frac{dy}{dx} \) and \( y = vx \) into equation (1):
\( v + x \frac{dv}{dx} = \frac{x^{2} - 2(vx)^{2} + x(vx)}{x^{2}} \)
\( v + x \frac{dv}{dx} = \frac{x^{2} - 2v^{2}x^{2} + vx^{2}}{x^{2}} \)
\( v + x \frac{dv}{dx} = \frac{x^{2}(1 - 2v^{2} + v)}{x^{2}} \)
\( v + x \frac{dv}{dx} = 1 - 2v^{2} + v \)
Now, separate the variables:
\( x \frac{dv}{dx} = 1 - 2v^{2} + v - v \)
\( x \frac{dv}{dx} = 1 - 2v^{2} \)
\( x \frac{dv}{dx} = -(2v^{2} - 1) \)
Now, we will separate the variables and integrate both sides:
\( \frac{dv}{2v^{2} - 1} = -\frac{dx}{x} \)
\( \int \frac{dv}{2v^{2} - 1} = -\int \frac{dx}{x} \)
\( \int \frac{dv}{2\left(v^{2} - \frac{1}{2}\right)} = -\int \frac{dx}{x} \)
\( \frac{1}{2} \int \frac{dv}{v^{2} - \left(\frac{1}{\sqrt{2}}\right)^{2}} = -\int \frac{dx}{x} \)
Using the integral formula \( \int \frac{1}{x^{2} - a^{2}} dx = \frac{1}{2a} \log\left|\frac{x - a}{x + a}\right| + C \):
\( \frac{1}{2} \cdot \frac{1}{2 \cdot \frac{1}{\sqrt{2}}} \log\left|\frac{v - \frac{1}{\sqrt{2}}}{v + \frac{1}{\sqrt{2}}}\right| = -\log|x| + C \)
\( \frac{1}{2\sqrt{2}} \log\left|\frac{\sqrt{2}v - 1}{\sqrt{2}v + 1}\right| = -\log|x| + C \)
Substitute back \( v = \frac{y}{x} \):
\( \frac{1}{2\sqrt{2}} \log\left|\frac{\sqrt{2}\frac{y}{x} - 1}{\sqrt{2}\frac{y}{x} + 1}\right| = -\log|x| + C \)
\( \frac{1}{2\sqrt{2}} \log\left|\frac{\frac{\sqrt{2}y - x}{x}}{\frac{\sqrt{2}y + x}{x}}\right| = -\log|x| + C \)
\( \frac{1}{2\sqrt{2}} \log\left|\frac{\sqrt{2}y - x}{\sqrt{2}y + x}\right| = -\log|x| + C \)
Therefore, the solution to the differential equation is \( \frac{1}{2\sqrt{2}} \log\left|\frac{\sqrt{2}y - x}{\sqrt{2}y + x}\right| = -\log|x| + C \).
In simple words: We changed the given equation to find \( \frac{dy}{dx} \). After confirming it was a homogeneous equation, we used \( y = vx \) to simplify it. Then, we separated the variables and performed integration using a special formula for \( \frac{1}{x^2 - a^2} \). Finally, we put the original variables back into the result.

Exam Tip: Be cautious when using the integral formula \( \int \frac{1}{x^{2} - a^{2}} dx \). Ensure you correctly identify \( x \) and \( a \) in your problem and manage the constant of integration carefully.

Question 6. \( xdy - ydx = \sqrt{x^{2}+y^{2}}dx \)
Answer: The given differential equation is:
\( xdy - ydx = \sqrt{x^{2}+y^{2}}dx \)
We can rearrange this equation to find \( \frac{dy}{dx} \):
\( xdy = ydx + \sqrt{x^{2}+y^{2}}dx \)
\( xdy = (y + \sqrt{x^{2}+y^{2}}) dx \)
\( \frac{dy}{dx} = \frac{y + \sqrt{x^{2}+y^{2}}}{x} \) ... (1)
Let's define \( F(x, y) = \frac{y + \sqrt{x^{2}+y^{2}}}{x} \).
Now, we will replace x with \( \lambda x \) and y with \( \lambda y \) in \( F(x, y) \):
\( F(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{(\lambda x)^{2}+(\lambda y)^{2}}}{\lambda x} \)
\( F(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{\lambda^{2}x^{2}+\lambda^{2}y^{2}}}{\lambda x} \)
\( F(\lambda x, \lambda y) = \frac{\lambda y + \lambda \sqrt{x^{2}+y^{2}}}{\lambda x} \)
\( F(\lambda x, \lambda y) = \frac{\lambda (y + \sqrt{x^{2}+y^{2}})}{\lambda x} \)
\( F(\lambda x, \lambda y) = \lambda^{0} \frac{y + \sqrt{x^{2}+y^{2}}}{x} \)
\( F(\lambda x, \lambda y) = \lambda^{0} F(x, y) \)
This shows that \( F(x, y) \) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve this, let's substitute \( y = vx \).
Differentiating \( y = vx \) with respect to x gives:
\( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Now, we will substitute this value of \( \frac{dy}{dx} \) and \( y = vx \) into equation (1):
\( v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^{2}+(vx)^{2}}}{x} \)
\( v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^{2}+v^{2}x^{2}}}{x} \)
\( v + x \frac{dv}{dx} = \frac{vx + x\sqrt{1+v^{2}}}{x} \)
\( v + x \frac{dv}{dx} = v + \sqrt{1+v^{2}} \)
Now, separate the variables:
\( x \frac{dv}{dx} = \sqrt{1+v^{2}} \)
Now, we will separate the variables and integrate both sides:
\( \frac{dv}{\sqrt{1+v^{2}}} = \frac{dx}{x} \)
\( \int \frac{dv}{\sqrt{1+v^{2}}} = \int \frac{dx}{x} \)
Using the integral formula \( \int \frac{1}{\sqrt{x^{2} + a^{2}}} dx = \log|x + \sqrt{x^{2} + a^{2}}| + C \):
\( \log|v + \sqrt{v^{2}+1}| = \log|x| + \log C \)
\( \log|v + \sqrt{v^{2}+1}| = \log|Cx| \)
\( v + \sqrt{v^{2}+1} = Cx \)
Substitute back \( v = \frac{y}{x} \):
\( \frac{y}{x} + \sqrt{\left(\frac{y}{x}\right)^{2}+1} = Cx \)
\( \frac{y}{x} + \sqrt{\frac{y^{2}}{x^{2}}+1} = Cx \)
\( \frac{y}{x} + \sqrt{\frac{y^{2}+x^{2}}{x^{2}}} = Cx \)
\( \frac{y}{x} + \frac{\sqrt{y^{2}+x^{2}}}{|x|} = Cx \)
Assuming \( x > 0 \):
\( \frac{y}{x} + \frac{\sqrt{y^{2}+x^{2}}}{x} = Cx \)
\( y + \sqrt{x^{2}+y^{2}} = Cx^{2} \)
Therefore, the solution to the differential equation is \( y + \sqrt{x^{2}+y^{2}} = Cx^{2} \).
In simple words: We first rearranged the equation to get \( \frac{dy}{dx} \). After checking that it was a homogeneous equation, we used the substitution \( y = vx \) to simplify it. We then separated the variables and integrated both sides using a specific formula for square roots. Finally, we substituted back the original variables to get the result.

Exam Tip: When dealing with square roots in the denominator for integration, remember the standard formulas for \( \int \frac{1}{\sqrt{x^2 \pm a^2}} dx \). The choice of constant as \( \log C \) simplifies the final logarithmic expression.

Question 7. \( \left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\} ydx = \left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\} xdy \)
Answer: The given differential equation is:
\( \left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\} ydx = \left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\} xdy \)
We can rearrange this equation to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{\left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\} y}{\left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\} x} \) ... (1)
Let's define \( F(x, y) = \frac{\left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\} y}{\left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\} x} \).
Now, we will replace x with \( \lambda x \) and y with \( \lambda y \) in \( F(x, y) \):
\( F(\lambda x, \lambda y) = \frac{\left\{\lambda x \cos\left(\frac{\lambda y}{\lambda x}\right) + \lambda y \sin\left(\frac{\lambda y}{\lambda x}\right)\right\} \lambda y}{\left\{\lambda y \sin\left(\frac{\lambda y}{\lambda x}\right) - \lambda x \cos\left(\frac{\lambda y}{\lambda x}\right)\right\} \lambda x} \)
\( F(\lambda x, \lambda y) = \frac{\left\{\lambda x \cos\left(\frac{y}{x}\right) + \lambda y \sin\left(\frac{y}{x}\right)\right\} \lambda y}{\left\{\lambda y \sin\left(\frac{y}{x}\right) - \lambda x \cos\left(\frac{y}{x}\right)\right\} \lambda x} \)
\( F(\lambda x, \lambda y) = \frac{\lambda^{2} \left\{x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right)\right\} y}{\lambda^{2} \left\{y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right)\right\} x} \)
\( F(\lambda x, \lambda y) = \lambda^{0} F(x, y) \)
This shows that \( F(x, y) \) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve this, let's substitute \( y = vx \).
Differentiating \( y = vx \) with respect to x gives:
\( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Now, we will substitute this value of \( \frac{dy}{dx} \) and \( y = vx \) into equation (1):
\( v + x \frac{dv}{dx} = \frac{\left\{x \cos(v) + vx \sin(v)\right\} vx}{\left\{vx \sin(v) - x \cos(v)\right\} x} \)
\( v + x \frac{dv}{dx} = \frac{x^{2}v \left\{\cos(v) + v \sin(v)\right\}}{x^{2} \left\{v \sin(v) - \cos(v)\right\}} \)
\( v + x \frac{dv}{dx} = \frac{v \cos(v) + v^{2} \sin(v)}{v \sin(v) - \cos(v)} \)
Now, separate the variables:
\( x \frac{dv}{dx} = \frac{v \cos(v) + v^{2} \sin(v)}{v \sin(v) - \cos(v)} - v \)
\( x \frac{dv}{dx} = \frac{v \cos(v) + v^{2} \sin(v) - v(v \sin(v) - \cos(v))}{v \sin(v) - \cos(v)} \)
\( x \frac{dv}{dx} = \frac{v \cos(v) + v^{2} \sin(v) - v^{2} \sin(v) + v \cos(v)}{v \sin(v) - \cos(v)} \)
\( x \frac{dv}{dx} = \frac{2v \cos(v)}{v \sin(v) - \cos(v)} \)
Now, we will separate the variables and integrate both sides:
\( \frac{v \sin(v) - \cos(v)}{2v \cos(v)} dv = \frac{dx}{x} \)
\( \frac{1}{2} \int \left(\frac{v \sin(v)}{v \cos(v)} - \frac{\cos(v)}{v \cos(v)}\right) dv = \int \frac{dx}{x} \)
\( \frac{1}{2} \int \left(\tan(v) - \frac{1}{v}\right) dv = \int \frac{dx}{x} \)
\( \frac{1}{2} (-\log|\cos v| - \log|v|) = \log|x| + \log C_{1} \)
\( - \frac{1}{2} (\log|\cos v| + \log|v|) = \log|x| + \log C_{1} \)
\( - \frac{1}{2} \log|v \cos v| = \log|x| + \log C_{1} \)
\( \log|v \cos v|^{-\frac{1}{2}} = \log|C_{1}x| \)
\( (v \cos v)^{-\frac{1}{2}} = C_{1}x \)
\( \frac{1}{\sqrt{v \cos v}} = C_{1}x \)
\( 1 = C_{1}x \sqrt{v \cos v} \)
Let \( C = \frac{1}{C_{1}} \).
\( \sqrt{v \cos v} = \frac{C}{x} \)
Square both sides:
\( v \cos v = \frac{C^{2}}{x^{2}} \)
Substitute back \( v = \frac{y}{x} \):
\( \frac{y}{x} \cos\left(\frac{y}{x}\right) = \frac{C^{2}}{x^{2}} \)
\( y \cos\left(\frac{y}{x}\right) = \frac{C^{2}}{x} \)
\( xy \cos\left(\frac{y}{x}\right) = C^{2} \)
Let \( K = C^{2} \).
\( xy \cos\left(\frac{y}{x}\right) = K \)
Therefore, the solution to the differential equation is \( xy \cos\left(\frac{y}{x}\right) = K \).
In simple words: This complex equation was first rearranged to find \( \frac{dy}{dx} \). After confirming it was homogeneous, we made the substitution \( y = vx \), which helped simplify the trigonometric terms. We then separated the variables and integrated both sides. Finally, we changed \( v \) back to \( \frac{y}{x} \) to obtain the final solution.

Exam Tip: Homogeneous equations with trigonometric functions often simplify nicely after \( y=vx \) substitution. Remember common integral formulas for \( \tan(v) \) and \( \frac{1}{v} \), and use logarithmic properties to combine constants.

Question 8. \( x \frac{dy}{dx} - y + x \sin\left(\frac{y}{x}\right) = 0 \)
Answer: The given differential equation is:
\( x \frac{dy}{dx} - y + x \sin\left(\frac{y}{x}\right) = 0 \)
We can rearrange this equation to find \( \frac{dy}{dx} \):
\( x \frac{dy}{dx} = y - x \sin\left(\frac{y}{x}\right) \)
\( \frac{dy}{dx} = \frac{y - x \sin\left(\frac{y}{x}\right)}{x} \) ... (1)
Let's define \( F(x, y) = \frac{y - x \sin\left(\frac{y}{x}\right)}{x} \).
Now, we will replace x with \( \lambda x \) and y with \( \lambda y \) in \( F(x, y) \):
\( F(\lambda x, \lambda y) = \frac{\lambda y - \lambda x \sin\left(\frac{\lambda y}{\lambda x}\right)}{\lambda x} \)
\( F(\lambda x, \lambda y) = \frac{\lambda y - \lambda x \sin\left(\frac{y}{x}\right)}{\lambda x} \)
\( F(\lambda x, \lambda y) = \frac{\lambda (y - x \sin\left(\frac{y}{x}\right))}{\lambda x} \)
\( F(\lambda x, \lambda y) = \lambda^{0} F(x, y) \)
This shows that \( F(x, y) \) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve this, let's substitute \( y = vx \).
Differentiating \( y = vx \) with respect to x gives:
\( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Now, we will substitute this value of \( \frac{dy}{dx} \) and \( y = vx \) into equation (1):
\( v + x \frac{dv}{dx} = \frac{vx - x \sin(v)}{x} \)
\( v + x \frac{dv}{dx} = v - \sin(v) \)
Now, separate the variables:
\( x \frac{dv}{dx} = -\sin(v) \)
Now, we will separate the variables and integrate both sides:
\( \frac{dv}{\sin(v)} = -\frac{dx}{x} \)
\( \int \csc(v) dv = -\int \frac{dx}{x} \)
\( \log|\csc(v) - \cot(v)| = -\log|x| + \log C \)
\( \log|\csc(v) - \cot(v)| = \log\left|\frac{C}{x}\right| \)
\( \csc(v) - \cot(v) = \frac{C}{x} \)
Substitute back \( v = \frac{y}{x} \):
\( \csc\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right) = \frac{C}{x} \)
We know that \( \csc\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right) = \frac{1}{\sin\left(\frac{y}{x}\right)} - \frac{\cos\left(\frac{y}{x}\right)}{\sin\left(\frac{y}{x}\right)} = \frac{1 - \cos\left(\frac{y}{x}\right)}{\sin\left(\frac{y}{x}\right)} \)
So, \( \frac{1 - \cos\left(\frac{y}{x}\right)}{\sin\left(\frac{y}{x}\right)} = \frac{C}{x} \)
\( x \left(1 - \cos\left(\frac{y}{x}\right)\right) = C \sin\left(\frac{y}{x}\right) \)
Therefore, the solution to the differential equation is \( x \left(1 - \cos\left(\frac{y}{x}\right)\right) = C \sin\left(\frac{y}{x}\right) \).
In simple words: We first rewrote the equation to get \( \frac{dy}{dx} \). After confirming it was a homogeneous equation, we used the substitution \( y = vx \). This simplified the equation greatly, allowing us to separate the variables and integrate. We used the integral for cosecant and then replaced \( v \) back with \( \frac{y}{x} \) to find the solution.

Exam Tip: The integral of \( \csc(v) \) is \( \log|\csc(v) - \cot(v)| \). Also, remember to simplify the trigonometric expression back to terms of \( y \) and \( x \) for the final answer.

Question 9. \( ydx + x \log\left(\frac{y}{x}\right)dy - 2x dy = 0 \)
Answer: The given differential equation is:
\( ydx + x \log\left(\frac{y}{x}\right)dy - 2x dy = 0 \)
We can rearrange this equation to find \( \frac{dy}{dx} \):
\( ydx = -(x \log\left(\frac{y}{x}\right) - 2x) dy \)
\( ydx = -x (\log\left(\frac{y}{x}\right) - 2) dy \)
\( \frac{dx}{dy} = -\frac{x}{y} (\log\left(\frac{y}{x}\right) - 2) \)
This equation is of the form \( \frac{dx}{dy} = F\left(\frac{x}{y}\right) \), which is also a homogeneous differential equation.
Let's define \( F(x, y) = -\frac{x (\log\left(\frac{y}{x}\right) - 2)}{y} \).
Now, we will replace x with \( \lambda x \) and y with \( \lambda y \) in \( F(x, y) \):
\( F(\lambda x, \lambda y) = -\frac{\lambda x (\log\left(\frac{\lambda y}{\lambda x}\right) - 2)}{\lambda y} \)
\( F(\lambda x, \lambda y) = -\frac{\lambda x (\log\left(\frac{y}{x}\right) - 2)}{\lambda y} \)
\( F(\lambda x, \lambda y) = \lambda^{0} F(x, y) \)
This shows that \( F(x, y) \) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve this, let's substitute \( x = vy \).
Differentiating \( x = vy \) with respect to y gives:
\( \frac{dx}{dy} = v + y \frac{dv}{dy} \)
Now, we will substitute this value of \( \frac{dx}{dy} \) and \( x = vy \) into the equation:
\( v + y \frac{dv}{dy} = -\frac{vy}{y} \left(\log\left(\frac{y}{vy}\right) - 2\right) \)
\( v + y \frac{dv}{dy} = -v \left(\log\left(\frac{1}{v}\right) - 2\right) \)
\( v + y \frac{dv}{dy} = -v (-\log v - 2) \)
\( v + y \frac{dv}{dy} = v \log v + 2v \)
Now, separate the variables:
\( y \frac{dv}{dy} = v \log v + 2v - v \)
\( y \frac{dv}{dy} = v \log v + v \)
\( y \frac{dv}{dy} = v (\log v + 1) \)
Now, we will separate the variables and integrate both sides:
\( \frac{dv}{v (\log v + 1)} = \frac{dy}{y} \)
\( \int \frac{dv}{v (\log v + 1)} = \int \frac{dy}{y} \)
For the left side, let \( u = \log v + 1 \), so \( du = \frac{1}{v} dv \).
\( \int \frac{du}{u} = \log|y| + \log C \)
\( \log|\log v + 1| = \log|Cy| \)
\( \log v + 1 = Cy \)
Substitute back \( v = \frac{x}{y} \):
\( \log\left(\frac{x}{y}\right) + 1 = Cy \)
Therefore, the solution to the differential equation is \( \log\left(\frac{x}{y}\right) + 1 = Cy \).
In simple words: We first rearranged the equation to get \( \frac{dx}{dy} \). After confirming it was a homogeneous equation (in terms of \( \frac{x}{y} \)), we used the substitution \( x = vy \). This simplified the equation, allowing us to separate the variables and integrate. We used a simple substitution for the integral and then replaced \( v \) back with \( \frac{x}{y} \) to find the final answer.

Exam Tip: When \( \frac{dy}{dx} \) is hard to isolate, check if \( \frac{dx}{dy} \) simplifies to a homogeneous form \( F(\frac{x}{y}) \). Then, substitute \( x=vy \). Remember that \( \log(\frac{1}{v}) = -\log v \).

Question 10. \( (1 + e^{\frac{x}{y}})dx + e^{\frac{x}{y}}(1 - \frac{x}{y})dy = 0 \)
Answer: The given differential equation is \( (1 + e^{\frac{x}{y}})dx + e^{\frac{x}{y}}(1 - \frac{x}{y})dy = 0 \). We can rewrite this to find \( \frac{dx}{dy} = \frac{-e^{\frac{x}{y}}(1 - \frac{x}{y})}{1+e^{\frac{x}{y}}} \). Let \( F(x, y) = \frac{-e^{\frac{x}{y}}(1 - \frac{x}{y})}{1+e^{\frac{x}{y}}} \). If we substitute \( x = \lambda x \) and \( y = \lambda y \), we get \( F(\lambda x, \lambda y) = \frac{-e^{\frac{x}{y}}(1 - \frac{x}{y})}{1+e^{\frac{x}{y}}} = \lambda^0 F(x, y) \). This result shows that \( F(x, y) \) is a homogeneous function of degree zero. Thus, the given differential equation is a homogeneous differential equation. We let \( x = vy \) and differentiate with respect to \( y \) to obtain \( \frac{dx}{dy} = v + y \frac{dv}{dy} \). Substituting these expressions into the equation, we get \( v + y \frac{dv}{dy} = \frac{-e^v (1 - v)}{1+e^v} \). After simplifying and rearranging the terms, we arrive at \( \frac{1+e^v}{v+e^v} dv = -\frac{dy}{y} \). Integrating both sides, we use the property \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| \) to get \( \log|v+e^v| = -\log|y| + \log c \). This can be written as \( \log|y(v+e^v)| = \log c \), which simplifies to \( y(v+e^v) = c \). Finally, substituting back \( v = \frac{x}{y} \), we obtain the general solution: \( x + y e^{\frac{x}{y}} = c \).
In simple words: We start by changing the given equation to find \( \frac{dx}{dy} \). We then check if it's a "homogeneous" equation, which means it behaves in a certain way when we scale x and y. Since it is, we replace \( x \) with \( vy \) and then solve it using integration. This involves separating the variables and performing the integration steps. After integrating and simplifying, we put \( x/y \) back in place of \( v \) to get the final solution.

Exam Tip: For homogeneous differential equations, remember to always verify the homogeneity by checking \( F(\lambda x, \lambda y) = \lambda^n F(x, y) \) before substituting \( y=vx \) or \( x=vy \). Choose \( x=vy \) if the equation is easier to express as \( \frac{dx}{dy} = F(x,y) \).

Solve Questions 11 to 15 to find the particular solution for each given differential equation under the given conditions.

Question 11. \( (x + y)dy + (x - y)dx = 0; \) જયારે \( x = 1 \) ત્યારે \( y = 1 \)
Answer: The given differential equation is \( (x + y)dy + (x - y)dx = 0 \). We rearrange it to get \( \frac{dy}{dx} = \frac{y-x}{x+y} \). We then let \( y = vx \) and differentiate it to obtain \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). Substituting these expressions into the equation, we get \( v + x \frac{dv}{dx} = \frac{vx - x}{x + vx} \). After simplifying and rearranging the terms, we arrive at \( \frac{1 + v}{1 + v^2} dv = -\frac{dx}{x} \). Integrating both sides gives us \( \arctan(v) + \frac{1}{2} \log|1 + v^2| = -\log|x| + c \). Now, we substitute \( v = \frac{y}{x} \) back into the solution, which yields \( \arctan(\frac{y}{x}) + \frac{1}{2} \log|x^2 + y^2| = c \). Using the initial condition that \( x = 1 \) when \( y = 1 \), we find the value of \( c \) to be \( \frac{\pi}{4} + \frac{1}{2} \log 2 \). Finally, substituting this value of \( c \) back into our general solution gives the particular solution: \( \arctan(\frac{y}{x}) + \frac{1}{2} \log|x^2 + y^2| = \frac{\pi}{4} + \frac{1}{2} \log 2 \). This particular solution can also be written as \( 2 \arctan(\frac{y}{x}) + \log(x^2 + y^2) = \frac{\pi}{2} + \log 2 \).
In simple words: We start by changing the equation into a form that's easier to solve. We assume \( y = vx \) and then replace it in the equation. This helps us separate the variables \( v \) and \( x \) so we can integrate both sides. After integration, we put \( y/x \) back for \( v \). Finally, we use the given starting values for \( x \) and \( y \) to find the specific constant, which gives us the particular answer.

Exam Tip: When solving for particular solutions, remember to substitute the initial conditions (x and y values) *after* finding the general solution to determine the constant of integration, \( c \). Make sure to simplify the logarithmic terms correctly.

Question 12. \( x^2 dy + (xy + y^2)dx = 0; \) જયારે \( x = 1 \) ત્યારે \( y = 1 \)
Answer: The given differential equation is \( x^2 dy + (xy + y^2)dx = 0 \). We rearrange it to get \( \frac{dy}{dx} = -\frac{xy + y^2}{x^2} \). We assume \( y = vx \) and then differentiate it to find \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). Substituting these into the equation, we get \( v + x \frac{dv}{dx} = -\frac{x(vx) + (vx)^2}{x^2} \). After simplifying and isolating variables, we get \( \frac{dv}{v(2 + v)} = -\frac{dx}{x} \). To integrate the left side, we use partial fractions, splitting \( \frac{1}{v(2+v)} \) into \( \frac{1}{2v} - \frac{1}{2(2+v)} \). Integrating both sides yields \( \frac{1}{2} \log|\frac{v}{2+v}| = -\log|x| + c \). We replace \( v \) with \( \frac{y}{x} \) to get \( \frac{1}{2} \log|\frac{y/x}{2+y/x}| = -\log|x| + c \). This simplifies to \( \frac{1}{2} \log|\frac{y}{2x+y}| = -\log|x| + c \). Multiplying by 2 and moving terms, we obtain \( \log|\frac{y}{2x+y}| = \log|\frac{C_1}{x^2}| \), which means \( \frac{y}{2x+y} = \frac{C_1}{x^2} \). This provides the general solution \( x^2 y = C_1(2x+y) \). Using the initial condition that \( x = 1 \) when \( y = 1 \), we find \( C_1 = \frac{1}{3} \). Substituting this value back, the particular solution is \( 3x^2 y = 2x+y \).
In simple words: We begin by rewriting the given equation to find \( \frac{dy}{dx} \). We then let \( y = vx \) and plug this into the equation. This helps us separate the \( v \) and \( x \) terms so we can integrate them separately. We use a method called partial fractions for the \( v \) part. After integrating, we substitute \( y/x \) back for \( v \). Finally, we use the given starting values of \( x=1 \) and \( y=1 \) to find the specific constant, which then gives us the exact solution for this particular case.

Exam Tip: Partial fractions are crucial for integrating rational functions. Remember to correctly determine the constants A and B before integration. Always substitute the initial conditions accurately to find the particular solution.

Question 13. \( [x \sin^2(\frac{y}{x}) - y]dx + x dy = 0; \) જયારે \( x = 1 \) ત્યારે \( y = \frac{\pi}{4} \)
Answer: The given differential equation is \( [x \sin^2(\frac{y}{x}) - y]dx + x dy = 0 \). We rearrange this to get \( \frac{dy}{dx} = \frac{y - x \sin^2(\frac{y}{x})}{x} \). We use the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). Plugging these into our equation, we obtain \( v + x \frac{dv}{dx} = \frac{vx - x \sin^2 v}{x} \). After simplifying, this becomes \( x \frac{dv}{dx} = -\sin^2 v \). We then separate the variables to get \( \csc^2 v dv = -\frac{dx}{x} \). Integrating both sides gives us \( -\cot v = -\log|x| + c' \), which can be written as \( \cot v = \log|x| + c \). Substituting back \( v = \frac{y}{x} \), we have the general solution: \( \cot(\frac{y}{x}) = \log|x| + c \). Using the initial condition that \( x = 1 \) when \( y = \frac{\pi}{4} \), we find \( c = 1 \). So, the particular solution is \( \cot(\frac{y}{x}) = \log|x| + 1 \).
In simple words: We take the given equation and reorder it to find \( \frac{dy}{dx} \). Then, we use the substitution \( y = vx \) to simplify the equation, making it easier to separate the terms with \( v \) from the terms with \( x \). We integrate both sides of the new equation. After integrating, we replace \( v \) with \( y/x \). Finally, we use the specific starting values for \( x \) and \( y \) to find the exact number for \( c \), giving us the specific solution.

Exam Tip: Watch out for trigonometric identities like \( \frac{1}{\sin^2 v} = \csc^2 v \). Integrating \( \csc^2 v \) gives \( -\cot v \). Remember \( \log 1 = 0 \).

Question 14. \( \frac{dy}{dx} - \frac{y}{x} + \csc (\frac{y}{x}) = 0; \) જયારે \( x = 1 \) ત્યારે \( y = 1 \)
Answer: The given differential equation is \( \frac{dy}{dx} - \frac{y}{x} + \csc(\frac{y}{x}) = 0 \). We rearrange it to get \( \frac{dy}{dx} = \frac{y}{x} - \csc(\frac{y}{x}) \). We use the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). Plugging these into our equation, we find \( v + x \frac{dv}{dx} = v - \csc v \). After simplifying, this becomes \( x \frac{dv}{dx} = -\csc v \). We then separate the variables to get \( \sin v dv = -\frac{dx}{x} \). Integrating both sides yields \( -\cos v = -\log|x| + c' \), which can be expressed as \( \cos v = \log|x| + c \). Substituting back \( v = \frac{y}{x} \), we obtain the general solution: \( \cos(\frac{y}{x}) = \log|x| + c \). Using the initial condition from the solution steps that \( x = 1 \) when \( y = 0 \), we find \( c = 1 \). So, the particular solution is \( \cos(\frac{y}{x}) = \log|x| + 1 \). This can also be written as \( \cos(\frac{y}{x}) = \log|e \cdot x| \).
In simple words: We begin by rewriting the given equation to isolate \( \frac{dy}{dx} \). We then substitute \( y = vx \) to simplify the equation, allowing us to separate the \( v \) and \( x \) terms. We integrate both sides. After integrating, we replace \( v \) with \( y/x \). Finally, we use the specific starting values for \( x \) and \( y \) given in the solution to find the exact value of the constant \( c \), which gives us the particular answer.

Exam Tip: Be careful with signs during integration, especially for trigonometric functions. The integral of \( \sin v \) is \( -\cos v \). Also, remember that \( \log|x| + 1 \) can be combined as \( \log|x| + \log e = \log|ex| \).

Question 15. \( 2xy + y^2 - 2x^2\frac{dy}{dx} = 0; \) જયારે \( x = 1 \) ત્યારે \( y = 2 \)
Answer: The given differential equation is \( 2xy + y^2 - 2x^2\frac{dy}{dx} = 0 \). We rearrange it to get \( \frac{dy}{dx} = \frac{2xy + y^2}{2x^2} \). We use the substitution \( y = vx \), which means \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). Plugging these into our equation, we find \( v + x \frac{dv}{dx} = \frac{2x(vx) + (vx)^2}{2x^2} \). After simplifying, this becomes \( x \frac{dv}{dx} = \frac{v^2}{2} \). We then separate the variables to get \( \frac{2}{v^2} dv = \frac{dx}{x} \). Integrating both sides yields \( -\frac{2}{v} = \log|x| + c \). Substituting back \( v = \frac{y}{x} \), we have the general solution: \( -\frac{2x}{y} = \log|x| + c \). Using the initial condition that \( x = 1 \) when \( y = 2 \), we find \( c = -1 \). So, the particular solution is \( -\frac{2x}{y} = \log|x| - 1 \), which can be written as \( y = \frac{2x}{1 - \log|x|} \).
In simple words: We start by rearranging the equation to solve for \( \frac{dy}{dx} \). We then use the substitution \( y = vx \) to simplify the equation, allowing us to separate the \( v \) and \( x \) terms. We integrate both sides. After integrating, we replace \( v \) with \( y/x \). Finally, we use the specific starting values for \( x \) and \( y \) to find the exact value of the constant \( c \), which gives us the particular answer.

Exam Tip: Pay attention to the power rule for integration, \( \int x^n dx = \frac{x^{n+1}}{n+1} \). Remember that \( \int v^{-2} dv = -v^{-1} \). Always double-check calculations when substituting initial conditions.

For Questions 16 And 17, Choose The Correct Option From The Given Alternatives To Make The Statement True:

Question 16. \( h(\frac{x}{y}) \) પ્રકારના સમપરિમાણ વિકલ સમીકરણનો ઉકેલ કયા આદેશ દ્વારા મેળવી શકાય ?
(A) y - vx
(B) v = yx
(C) x = vy
(D) x = v
Answer: (C) x = vy
In simple words: When the differential equation is given in the form \( \frac{dx}{dy} = h(\frac{x}{y}) \), the best way to solve it is by replacing \( x \) with \( vy \). This substitution helps simplify the equation so it can be solved.

Exam Tip: If the differential equation is of the form \( \frac{dy}{dx} = f(\frac{y}{x}) \), use \( y = vx \). If it's \( \frac{dx}{dy} = f(\frac{x}{y}) \), use \( x = vy \). This choice simplifies the separation of variables.

Question 17. નીચેનામાંથી કયું વિકલ સમીકરણ સમપરિમાણ છે ?
(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) y² dx + (x² - xy - y²) dy = 0
Answer: (D) y² dx + (x² - xy - y²) dy = 0
In simple words: A homogeneous differential equation is one where, if you replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \), all the \( \lambda \) terms cancel out, leaving the original function multiplied by \( \lambda \) to some power (usually zero). Checking each option: (A) has constants that don't cancel. (B) has different powers of \( \lambda \) in the top and bottom. (C) has mixed powers of \( \lambda \) in the numerator. Only (D) simplifies correctly to be homogeneous, where all \( \lambda \) terms cancel out, meaning it's homogeneous of degree zero.

Exam Tip: To check if a differential equation \( M(x,y)dx + N(x,y)dy = 0 \) is homogeneous, both \( M(x,y) \) and \( N(x,y) \) must be homogeneous functions of the same degree. Alternatively, rewrite it as \( \frac{dy}{dx} = F(x,y) \) and check if \( F(\lambda x, \lambda y) = \lambda^0 F(x,y) \). Constant terms or different degrees in different parts of \( F(x,y) \) indicate non-homogeneity.