RBSE Solutions Class 12 Maths Chapter 12 Differential Equation Exercise 12.6

Solve the following differential equations:

Question 1. \( x^2 ydx - (x^3 + y^3) dy = 0 \)
Answer: We begin by rewriting the given differential equation in the form \( \frac{dy}{dx} = f(x, y) \).
From the equation \( x^2 ydx - (x^3 + y^3) dy = 0 \), we move terms to isolate \( \frac{dy}{dx} \):
\( x^2 ydx = (x^3 + y^3) dy \)
\( \implies \frac{dy}{dx} = \frac{x^2 y}{x^3 + y^3} \) ...(1)
Next, we check if this is a homogeneous equation by substituting \( y = vx \), which implies \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
So, \( v + x \frac{dv}{dx} = \frac{x^2 (vx)}{x^3 + (vx)^3} \)
\( \implies v + x \frac{dv}{dx} = \frac{vx^3}{x^3 + v^3 x^3} \)
\( \implies v + x \frac{dv}{dx} = \frac{vx^3}{x^3 (1 + v^3)} \)
\( \implies v + x \frac{dv}{dx} = \frac{v}{1 + v^3} \)
Now, we separate the variables:
\( x \frac{dv}{dx} = \frac{v}{1 + v^3} - v \)
\( \implies x \frac{dv}{dx} = \frac{v - v(1 + v^3)}{1 + v^3} \)
\( \implies x \frac{dv}{dx} = \frac{v - v - v^4}{1 + v^3} \)
\( \implies x \frac{dv}{dx} = \frac{-v^4}{1 + v^3} \)
\( \implies \frac{1 + v^3}{-v^4} dv = \frac{1}{x} dx \)
\( \implies - \left( \frac{1}{v^4} + \frac{v^3}{v^4} \right) dv = \frac{1}{x} dx \)
\( \implies - \left( v^{-4} + \frac{1}{v} \right) dv = \frac{1}{x} dx \)
Integrate both sides:
\( - \int \left( v^{-4} + \frac{1}{v} \right) dv = \int \frac{1}{x} dx \)
\( \implies - \left( \frac{v^{-3}}{-3} + \log|v| \right) = \log|x| + \log|C_1| \)
\( \implies \frac{1}{3v^3} - \log|v| = \log|x| + \log|C_1| \)
\( \implies \frac{1}{3v^3} = \log|x| + \log|v| + \log|C_1| \)
\( \implies \frac{1}{3v^3} = \log(C_1vx) \)
Finally, substitute back \( v = \frac{y}{x} \):
\( \frac{1}{3 \left(\frac{y}{x}\right)^3} = \log\left(C_1 \frac{y}{x} x\right) \)
\( \implies \frac{x^3}{3y^3} = \log(C_1 y) \)
We can also write this as \( \log(C_1 y) = \frac{x^3}{3y^3} \). This type of equation is often solved by assuming \( y = vx \) to simplify the structure.
In simple words: First, we rearranged the equation to get \( \frac{dy}{dx} \). Then, we assumed \( y \) is a multiple of \( x \) (called \( vx \)) to make the equation simpler. After some algebra, we separated the \( v \) terms and the \( x \) terms so we could integrate both sides. Finally, we put \( y/x \) back in place of \( v \) to get the solution.

🎯 Exam Tip: When solving homogeneous differential equations, always remember to substitute back \( v = \frac{y}{x} \) at the very end to express the solution in terms of the original variables \( x \) and \( y \).

Question 2. \( \frac{dy}{dx} = \frac{y}{x} + \sin\left(\frac{y}{x}\right) \)
Answer: The given differential equation is \( \frac{dy}{dx} = \frac{y}{x} + \sin\left(\frac{y}{x}\right) \) ...(1)
This equation is already in the form \( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \), which means it is a homogeneous differential equation of zero degree.
To solve it, we put \( y = vx \).
Differentiating \( y = vx \) with respect to \( x \), we get \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) ...(2)
Substitute (2) into (1):
\( v + x \frac{dv}{dx} = v + \sin(v) \)
\( \implies x \frac{dv}{dx} = \sin(v) \)
Now, separate the variables:
\( \frac{dv}{\sin(v)} = \frac{dx}{x} \)
\( \implies \operatorname{cosec}(v) dv = \frac{dx}{x} \)
Integrate both sides:
\( \int \operatorname{cosec}(v) dv = \int \frac{dx}{x} \)
\( \implies \log|\operatorname{cosec}(v) - \cot(v)| = \log|x| + \log|c| \)
\( \implies \log|\operatorname{cosec}(v) - \cot(v)| = \log|cx| \)
Comparing both sides, we get:
\( \operatorname{cosec}(v) - \cot(v) = cx \)
Now, substitute back \( v = \frac{y}{x} \):
\( \operatorname{cosec}\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right) = cx \)
Using trigonometric identities, \( \operatorname{cosec}(v) = \frac{1}{\sin(v)} \) and \( \cot(v) = \frac{\cos(v)}{\sin(v)} \):
\( \frac{1}{\sin\left(\frac{y}{x}\right)} - \frac{\cos\left(\frac{y}{x}\right)}{\sin\left(\frac{y}{x}\right)} = cx \)
\( \implies \frac{1 - \cos\left(\frac{y}{x}\right)}{\sin\left(\frac{y}{x}\right)} = cx \)
The final answer represents the relationship between \( x \) and \( y \) that satisfies the given differential equation.
In simple words: The equation already looked like \( y/x \) was important. So we replaced \( y/x \) with \( v \) and then used a trick to change \( dy/dx \). This let us separate the parts with \( v \) from the parts with \( x \) and solve them separately using integration. Finally, we put \( y/x \) back where \( v \) was.

🎯 Exam Tip: Recognize that \( \operatorname{cosec}(v) - \cot(v) \) simplifies nicely, which is a common pattern in these types of problems. Remember the integration formulas for trigonometric functions.

Question 3. \( x \frac{dy}{dx} + \frac{y^2}{x} = y \)
Answer: The given differential equation is \( x \frac{dy}{dx} + \frac{y^2}{x} = y \).
First, we rewrite the equation to isolate \( \frac{dy}{dx} \):
\( x \frac{dy}{dx} = y - \frac{y^2}{x} \)
\( \implies x \frac{dy}{dx} = \frac{xy - y^2}{x} \)
\( \implies \frac{dy}{dx} = \frac{xy - y^2}{x^2} \) ...(1)
This is a homogeneous equation because each term in the numerator and denominator has the same degree (2).
We substitute \( y = vx \), which gives \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \frac{x(vx) - (vx)^2}{x^2} \)
\( \implies v + x \frac{dv}{dx} = \frac{vx^2 - v^2 x^2}{x^2} \)
\( \implies v + x \frac{dv}{dx} = v - v^2 \)
\( \implies x \frac{dv}{dx} = -v^2 \)
Now, separate the variables:
\( \frac{dv}{-v^2} = \frac{dx}{x} \)
\( \implies - \frac{1}{v^2} dv = \frac{1}{x} dx \)
Integrate both sides:
\( - \int v^{-2} dv = \int \frac{1}{x} dx \)
\( \implies - \left( \frac{v^{-1}}{-1} \right) = \log|x| + C \)
\( \implies \frac{1}{v} = \log|x| + C \)
Finally, substitute back \( v = \frac{y}{x} \):
\( \frac{1}{\frac{y}{x}} = \log|x| + C \)
\( \implies \frac{x}{y} = \log|x| + C \)
This can also be written as \( x = y(\log|x| + C) \). This method uses a substitution to turn a complicated equation into one that's easier to integrate.
In simple words: We first cleaned up the equation to get \( dy/dx \) alone. Since it was a homogeneous equation, we used \( y = vx \) to simplify it. After replacing \( y \) and \( dy/dx \), we managed to put all the \( v \) parts on one side and all the \( x \) parts on the other. Then we integrated both sides and put \( y/x \) back in the answer.

🎯 Exam Tip: Remember that \( \int v^{-2} dv = -v^{-1} \). Be careful with the signs during integration. Also, always check if the equation is homogeneous before attempting the \( y=vx \) substitution.

Question 4. \( x \sin\left(\frac{y}{x}\right) \frac{dy}{dx} = y \sin\left(\frac{y}{x}\right) - x \)
Answer: The given differential equation is \( x \sin\left(\frac{y}{x}\right) \frac{dy}{dx} = y \sin\left(\frac{y}{x}\right) - x \).
First, we rewrite the equation to isolate \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y \sin\left(\frac{y}{x}\right) - x}{x \sin\left(\frac{y}{x}\right)} \) ...(1)
Let \( F(x, y) = \frac{y \sin\left(\frac{y}{x}\right) - x}{x \sin\left(\frac{y}{x}\right)} \).
To check if it is homogeneous, substitute \( x = \lambda x \) and \( y = \lambda y \):
\( F(\lambda x, \lambda y) = \frac{\lambda y \sin\left(\frac{\lambda y}{\lambda x}\right) - \lambda x}{\lambda x \sin\left(\frac{\lambda y}{\lambda x}\right)} \)
\( \implies F(\lambda x, \lambda y) = \frac{\lambda \left( y \sin\left(\frac{y}{x}\right) - x \right)}{\lambda \left( x \sin\left(\frac{y}{x}\right) \right)} \)
\( \implies F(\lambda x, \lambda y) = \lambda^0 \frac{y \sin\left(\frac{y}{x}\right) - x}{x \sin\left(\frac{y}{x}\right)} \)
\( \implies F(\lambda x, \lambda y) = \lambda^0 F(x, y) \).
Thus, \( F(x, y) \) is a homogeneous function of zero degree. So, the given differential equation is homogeneous.
We use the substitution \( y = vx \), which implies \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \frac{vx \sin(v) - x}{x \sin(v)} \)
\( \implies v + x \frac{dv}{dx} = \frac{x(v \sin(v) - 1)}{x \sin(v)} \)
\( \implies v + x \frac{dv}{dx} = \frac{v \sin(v) - 1}{\sin(v)} \)
\( \implies x \frac{dv}{dx} = \frac{v \sin(v) - 1}{\sin(v)} - v \)
\( \implies x \frac{dv}{dx} = \frac{v \sin(v) - 1 - v \sin(v)}{\sin(v)} \)
\( \implies x \frac{dv}{dx} = \frac{-1}{\sin(v)} \)
Now, separate the variables:
\( \sin(v) dv = - \frac{1}{x} dx \)
Integrate both sides:
\( \int \sin(v) dv = - \int \frac{1}{x} dx \)
\( \implies -\cos(v) = -\log|x| + C_1 \)
\( \implies \cos(v) = \log|x| - C_1 \)
Let \( C = -C_1 \).
\( \implies \cos(v) = \log|x| + C \)
Finally, substitute back \( v = \frac{y}{x} \):
\( \cos\left(\frac{y}{x}\right) = \log|x| + C \)
This can also be written as \( \log|Cx| = \cos\left(\frac{y}{x}\right) \), which means \( Cx = e^{\cos(y/x)} \). A homogeneous equation simplifies easily with the substitution \( y=vx \).
In simple words: We first rearranged the equation for \( dy/dx \) and checked if it was a homogeneous equation (it was!). Then, we swapped \( y \) with \( vx \) and \( dy/dx \) with \( v + x dv/dx \). This allowed us to put all the \( v \) parts on one side and all the \( x \) parts on the other. After integrating, we replaced \( v \) with \( y/x \) to get the final answer.

🎯 Exam Tip: When integrating \( \sin(v) \), remember that it results in \( -\cos(v) \). Also, be careful with signs when rearranging the constant of integration.

Question 5. \( xdy - ydx = \sqrt{x^2 + y^2} dx \)
Answer: The given differential equation is \( xdy - ydx = \sqrt{x^2 + y^2} dx \).
First, we rearrange the equation to isolate \( \frac{dy}{dx} \):
\( xdy = ydx + \sqrt{x^2 + y^2} dx \)
\( \implies xdy = (y + \sqrt{x^2 + y^2}) dx \)
\( \implies \frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x} \) ...(1)
To check if it is homogeneous, let \( F(x, y) = \frac{y + \sqrt{x^2 + y^2}}{x} \).
Substitute \( x = \lambda x \) and \( y = \lambda y \):
\( F(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{(\lambda x)^2 + (\lambda y)^2}}{\lambda x} \)
\( \implies F(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{\lambda^2 x^2 + \lambda^2 y^2}}{\lambda x} \)
\( \implies F(\lambda x, \lambda y) = \frac{\lambda y + \lambda \sqrt{x^2 + y^2}}{\lambda x} \)
\( \implies F(\lambda x, \lambda y) = \frac{\lambda (y + \sqrt{x^2 + y^2})}{\lambda x} \)
\( \implies F(\lambda x, \lambda y) = \lambda^0 F(x, y) \).
Thus, \( F(x, y) \) is a homogeneous function of zero degree. So, the given differential equation is homogeneous.
We use the substitution \( y = vx \), which implies \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + (vx)^2}}{x} \)
\( \implies v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2 x^2}}{x} \)
\( \implies v + x \frac{dv}{dx} = \frac{vx + x\sqrt{1 + v^2}}{x} \)
\( \implies v + x \frac{dv}{dx} = v + \sqrt{1 + v^2} \)
\( \implies x \frac{dv}{dx} = \sqrt{1 + v^2} \)
Now, separate the variables:
\( \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x} \)
Integrate both sides:
\( \int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x} \)
The integral of \( \frac{1}{\sqrt{1 + v^2}} \) is \( \log|v + \sqrt{1 + v^2}| \).
So, \( \log|v + \sqrt{1 + v^2}| = \log|x| + \log|C| \)
\( \implies \log|v + \sqrt{1 + v^2}| = \log|Cx| \)
Comparing both sides, we get:
\( v + \sqrt{1 + v^2} = Cx \)
Finally, substitute back \( v = \frac{y}{x} \):
\( \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} = Cx \)
\( \implies \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = Cx \)
\( \implies \frac{y}{x} + \sqrt{\frac{x^2 + y^2}{x^2}} = Cx \)
\( \implies \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{|x|} = Cx \)
Assuming \( x > 0 \), we have:
\( \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = Cx \)
\( \implies y + \sqrt{x^2 + y^2} = Cx^2 \)
This solution shows how to handle square roots in homogeneous equations by simplifying inside the radical.
In simple words: We first organized the equation to get \( dy/dx \) by itself. We confirmed it was a homogeneous equation. Then, we used the substitution \( y = vx \) to simplify the problem, separating the \( v \) and \( x \) terms. After integrating both sides, we replaced \( v \) with \( y/x \) to get the final answer in terms of \( x \) and \( y \).

🎯 Exam Tip: Remember the integral formula for \( \frac{1}{\sqrt{a^2 + x^2}} \). When dealing with \( \sqrt{x^2} \), it simplifies to \( |x| \), so be mindful of positive or negative \( x \) values, usually assuming \( x>0 \) for simplicity.

Question 6. \( (x^2 + y^2) dx = 2xydy \)
Answer: The given differential equation is \( (x^2 + y^2) dx = 2xydy \).
First, we rearrange the equation to isolate \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \) ...(1)
This is a homogeneous differential equation because each term ( \( x^2, y^2, 2xy \) ) has degree 2.
We use the substitution \( y = vx \), which implies \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} \)
\( \implies v + x \frac{dv}{dx} = \frac{x^2 + v^2 x^2}{2v x^2} \)
\( \implies v + x \frac{dv}{dx} = \frac{x^2(1 + v^2)}{2v x^2} \)
\( \implies v + x \frac{dv}{dx} = \frac{1 + v^2}{2v} \)
Now, separate the variables:
\( x \frac{dv}{dx} = \frac{1 + v^2}{2v} - v \)
\( \implies x \frac{dv}{dx} = \frac{1 + v^2 - 2v^2}{2v} \)
\( \implies x \frac{dv}{dx} = \frac{1 - v^2}{2v} \)
\( \implies \frac{2v}{1 - v^2} dv = \frac{1}{x} dx \)
Integrate both sides:
\( \int \frac{2v}{1 - v^2} dv = \int \frac{1}{x} dx \)
For the left side, let \( u = 1 - v^2 \), so \( du = -2v dv \). This means \( 2v dv = -du \).
\( \int \frac{-du}{u} = \int \frac{1}{x} dx \)
\( \implies -\log|u| = \log|x| + \log|C| \)
\( \implies -\log|1 - v^2| = \log|x| + \log|C| \)
\( \implies -\log|1 - v^2| = \log|Cx| \)
\( \implies \log|(1 - v^2)^{-1}| = \log|Cx| \)
\( \implies \frac{1}{1 - v^2} = Cx \)
\( \implies 1 = Cx(1 - v^2) \)
Finally, substitute back \( v = \frac{y}{x} \):
\( 1 = Cx \left(1 - \left(\frac{y}{x}\right)^2 \right) \)
\( \implies 1 = Cx \left(1 - \frac{y^2}{x^2} \right) \)
\( \implies 1 = Cx \left(\frac{x^2 - y^2}{x^2} \right) \)
\( \implies 1 = C \frac{x^2 - y^2}{x} \)
\( \implies x = C(x^2 - y^2) \)
This result confirms that careful substitution and integration can lead to a simplified form.
In simple words: First, we made the equation ready by getting \( dy/dx \) alone. We saw it was a homogeneous equation, so we used the special substitution \( y = vx \). This helped us split the equation into parts with \( v \) and parts with \( x \). We then integrated each side and put \( y/x \) back to get the final answer.

🎯 Exam Tip: When integrating \( \frac{2v}{1 - v^2} \), recognize that the numerator is almost the derivative of the denominator (with a sign difference), which suggests a simple logarithmic integral using substitution.

Question 7. \( \left(1+e^{\frac{x}{y}}\right) dx + e^{\frac{x}{y}} \left(1-\frac{x}{y}\right) dy = 0 \)
Answer: The given differential equation is \( \left(1+e^{\frac{x}{y}}\right) dx + e^{\frac{x}{y}} \left(1-\frac{x}{y}\right) dy = 0 \).
Since the terms involve \( \frac{x}{y} \), it's easier to express this equation in terms of \( \frac{dx}{dy} \).
\( \left(1+e^{\frac{x}{y}}\right) dx = - e^{\frac{x}{y}} \left(1-\frac{x}{y}\right) dy \)
\( \implies \frac{dx}{dy} = - \frac{e^{\frac{x}{y}} \left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}} \) ...(1)
This is a homogeneous differential equation because it can be written as \( \frac{dx}{dy} = f\left(\frac{x}{y}\right) \).
For this type of equation, we use the substitution \( x = vy \).
Differentiating \( x = vy \) with respect to \( y \), we get \( \frac{dx}{dy} = v + y \frac{dv}{dy} \) ...(2)
Substitute (2) into (1):
\( v + y \frac{dv}{dy} = - \frac{e^{v} (1-v)}{1+e^{v}} \)
Now, separate the variables:
\( y \frac{dv}{dy} = - \frac{e^{v} (1-v)}{1+e^{v}} - v \)
\( \implies y \frac{dv}{dy} = - \frac{e^{v} - ve^{v}}{1+e^{v}} - v \)
\( \implies y \frac{dv}{dy} = \frac{-e^{v} + ve^{v} - v(1+e^{v})}{1+e^{v}} \)
\( \implies y \frac{dv}{dy} = \frac{-e^{v} + ve^{v} - v - ve^{v}}{1+e^{v}} \)
\( \implies y \frac{dv}{dy} = \frac{-e^{v} - v}{1+e^{v}} \)
\( \implies y \frac{dv}{dy} = - \frac{v+e^{v}}{1+e^{v}} \)
\( \implies \frac{1+e^{v}}{v+e^{v}} dv = - \frac{1}{y} dy \)
Integrate both sides:
\( \int \frac{1+e^{v}}{v+e^{v}} dv = - \int \frac{1}{y} dy \)
For the left side, let \( t = v+e^{v} \), so \( dt = (1+e^{v}) dv \).
\( \int \frac{dt}{t} = - \int \frac{1}{y} dy \)
\( \implies \log|t| = -\log|y| + \log|C| \)
\( \implies \log|v+e^{v}| = -\log|y| + \log|C| \)
\( \implies \log|v+e^{v}| = \log\left|\frac{C}{y}\right| \)
Comparing both sides, we get:
\( v+e^{v} = \frac{C}{y} \)
Finally, substitute back \( v = \frac{x}{y} \):
\( \frac{x}{y} + e^{\frac{x}{y}} = \frac{C}{y} \)
Multiply by \( y \) to clear the denominator:
\( x + y e^{\frac{x}{y}} = C \)
This equation demonstrates how to adapt the homogeneous substitution when the ratio is \( x/y \) instead of \( y/x \).
In simple words: Because the equation used \( x/y \) a lot, we changed it to find \( dx/dy \) and then used \( x = vy \) for our substitution. After doing this, we separated the \( v \) and \( y \) parts to integrate them. Finally, we put \( x/y \) back in place of \( v \) to get the solution.

🎯 Exam Tip: If the differential equation involves the ratio \( \frac{x}{y} \) instead of \( \frac{y}{x} \), it is generally more efficient to use the substitution \( x=vy \) and differentiate with respect to \( y \) to find \( \frac{dx}{dy} \).

Question 8. \( (3xy + y^2) dx + (x^2 + xy) dy = 0 \)
Answer: The given differential equation is \( (3xy + y^2) dx + (x^2 + xy) dy = 0 \).
First, we rewrite the equation to isolate \( \frac{dy}{dx} \):
\( (x^2 + xy) dy = - (3xy + y^2) dx \)
\( \implies \frac{dy}{dx} = - \frac{3xy + y^2}{x^2 + xy} \) ...(1)
This is a homogeneous differential equation because each term in the numerator ( \( 3xy, y^2 \) ) and denominator ( \( x^2, xy \) ) has degree 2.
We use the substitution \( y = vx \), which implies \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
Substitute these into equation (1):
\( v + x \frac{dv}{dx} = - \frac{3x(vx) + (vx)^2}{x^2 + x(vx)} \)
\( \implies v + x \frac{dv}{dx} = - \frac{3vx^2 + v^2 x^2}{x^2 + vx^2} \)
\( \implies v + x \frac{dv}{dx} = - \frac{x^2(3v + v^2)}{x^2(1 + v)} \)
\( \implies v + x \frac{dv}{dx} = - \frac{3v + v^2}{1 + v} \)
Now, separate the variables:
\( x \frac{dv}{dx} = - \frac{3v + v^2}{1 + v} - v \)
\( \implies x \frac{dv}{dx} = \frac{-(3v + v^2) - v(1 + v)}{1 + v} \)
\( \implies x \frac{dv}{dx} = \frac{-3v - v^2 - v - v^2}{1 + v} \)
\( \implies x \frac{dv}{dx} = \frac{-4v - 2v^2}{1 + v} \)
\( \implies x \frac{dv}{dx} = \frac{-2v(2 + v)}{1 + v} \)
This completes the separation of variables, ready for integration in the next step.
In simple words: We first isolated \( dy/dx \) from the given equation. We recognized it as a homogeneous equation because all parts had the same overall power. By replacing \( y \) with \( vx \) and \( dy/dx \) with its equivalent, we simplified the equation. This allowed us to separate the \( v \) terms from the \( x \) terms, preparing for integration.

🎯 Exam Tip: After substituting \( y=vx \), ensure you correctly expand and simplify all terms, especially when combining fractions, to avoid errors in variable separation.

Question 9. \( x^2 \frac{dy}{dx} = x^2 + xy + y^2 \)
Answer: First, we rewrite the given differential equation by dividing by \( x^2 \) to express it in terms of \( \frac{dy}{dx} \).
\( \frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} \)
\( \frac{dy}{dx} = 1 + \frac{y}{x} + \left( \frac{y}{x} \right)^2 \)
This equation is homogeneous because all terms have the same degree when considering \( x \) and \( y \). To solve it, we make the substitution \( y = vx \).
\( \implies \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Substitute this into the differential equation:
\( v + x \frac{dv}{dx} = 1 + v + v^2 \)
Subtract \( v \) from both sides:
\( x \frac{dv}{dx} = 1 + v^2 \)
Now, we separate the variables, putting all \( v \) terms on one side and all \( x \) terms on the other:
\( \frac{dv}{1+v^2} = \frac{dx}{x} \)
Integrate both sides:
\( \int \frac{dv}{1+v^2} = \int \frac{dx}{x} \)
\( \tan^{-1} v = \log |x| + C \)
Finally, substitute back \( v = \frac{y}{x} \) to get the solution in terms of \( x \) and \( y \). This substitution helps transform the equation into a separable form, making it solvable.
\( \tan^{-1} \left( \frac{y}{x} \right) = \log |x| + C \)
In simple words: To solve this equation, first make sure it's "homogeneous" (meaning all parts have the same total power). Then, replace \( y \) with \( vx \) and \( \frac{dy}{dx} \) with \( v + x \frac{dv}{dx} \). This lets you separate the \( v \) and \( x \) parts, integrate them, and then change \( v \) back to \( \frac{y}{x} \) to find the final answer.

🎯 Exam Tip: Always check if a differential equation is homogeneous before applying the \( y = vx \) substitution. Remember the standard integral for \( \frac{1}{1+v^2} \) and \( \frac{1}{x} \).

Question 10. \( x(x-y) dy = y(x+y) dx \)
Answer: First, we rearrange the given differential equation to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y(x+y)}{x(x-y)} \)
Divide the numerator and denominator by \( x^2 \) to check for homogeneity:
\( \frac{dy}{dx} = \frac{\frac{y}{x}(1+\frac{y}{x})}{1-\frac{y}{x}} \)
Since the equation can be expressed as a function of \( \frac{y}{x} \), it is a homogeneous differential equation. Now, we use the substitution \( y = vx \).
\( \implies \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into the equation:
\( v + x \frac{dv}{dx} = \frac{vx(x+vx)}{x(x-vx)} \)
\( v + x \frac{dv}{dx} = \frac{vx^2(1+v)}{x^2(1-v)} \)
\( v + x \frac{dv}{dx} = \frac{v(1+v)}{1-v} \)
Now, isolate \( x \frac{dv}{dx} \):
\( x \frac{dv}{dx} = \frac{v(1+v)}{1-v} - v \)
\( x \frac{dv}{dx} = \frac{v(1+v) - v(1-v)}{1-v} \)
\( x \frac{dv}{dx} = \frac{v+v^2-v+v^2}{1-v} \)
\( x \frac{dv}{dx} = \frac{2v^2}{1-v} \)
Separate the variables, moving all \( v \) terms to one side and \( x \) terms to the other:
\( \frac{1-v}{2v^2} dv = \frac{dx}{x} \)
Rewrite the left side:
\( \left( \frac{1}{2v^2} - \frac{v}{2v^2} \right) dv = \frac{dx}{x} \)
\( \left( \frac{1}{2v^2} - \frac{1}{2v} \right) dv = \frac{dx}{x} \)
Integrate both sides:
\( \int \left( \frac{1}{2v^2} - \frac{1}{2v} \right) dv = \int \frac{dx}{x} \)
\( \frac{1}{2} \int v^{-2} dv - \frac{1}{2} \int \frac{1}{v} dv = \int \frac{1}{x} dx \)
\( \frac{1}{2} \left( \frac{v^{-1}}{-1} \right) - \frac{1}{2} \log |v| = \log |x| + C \)
\( -\frac{1}{2v} - \frac{1}{2} \log |v| = \log |x| + C \)
Multiply the entire equation by 2:
\( -\frac{1}{v} - \log |v| = 2 \log |x| + 2C \)
Let \( 2C = C_1 \) (another arbitrary constant). Also, use \( 2 \log |x| = \log (x^2) \):
\( -\frac{1}{v} - \log |v| = \log (x^2) + C_1 \)
Finally, substitute back \( v = \frac{y}{x} \) to get the solution in terms of \( x \) and \( y \). This transformation makes complex equations solvable by isolating variables for integration.
\( -\frac{x}{y} - \log \left| \frac{y}{x} \right| = \log (x^2) + C_1 \)
Using properties of logarithms \( \log \left| \frac{y}{x} \right| = \log |y| - \log |x| \):
\( -\frac{x}{y} - (\log |y| - \log |x|) = \log (x^2) + C_1 \)
\( -\frac{x}{y} - \log |y| + \log |x| = 2 \log |x| + C_1 \)
Rearranging terms:
\( -\frac{x}{y} - \log |y| - \log |x| = C_1 \)
\( -\frac{x}{y} - (\log |y| + \log |x|) = C_1 \)
\( -\frac{x}{y} - \log |xy| = C_1 \)
We can write this as \( \frac{x}{y} + \log |xy| + C_1 = 0 \).
In simple words: First, rearrange the equation to find \( \frac{dy}{dx} \). Since it's a homogeneous equation, replace \( y \) with \( vx \) and simplify. Then, move all terms with \( v \) to one side and all terms with \( x \) to the other. Integrate both sides, remembering to split fractions if needed. Finally, replace \( v \) back with \( \frac{y}{x} \) to get the solution.

🎯 Exam Tip: Pay close attention to algebraic simplifications and logarithmic properties during integration and substitution. Mistakes in these steps are common pitfalls.