Question

The general solution of the differential equation \( (x-y)dy=(x+y)dx \) is

• A. \( \tan^{-1}\left(\frac{y}{x}\right)=c\sqrt{x^2+y^2} \)
• B. \( \tan^{-1}\left(\frac{y}{x}\right)=x^2+y^2+c \)
• C. \( e^{\tan^{-1}\left(\frac{y}{x}\right)}=\dfrac{c\sqrt{x^2+y^2}}{x} \)
• D. \( e^{\tan^{-1}\left(\frac{y}{x}\right)}=c\sqrt{x^2+y^2} \)

Hint:

For homogeneous differential equations, use the substitution \(y=vx\), then separate variables and integrate.

Answer 1

The Correct Option is D

Step 1: Write the given differential equation.
The given differential equation is:
\[ (x-y)dy=(x+y)dx. \]
Dividing both sides by \(dx\), we get:
\[ (x-y)\frac{dy}{dx}=x+y. \]

Step 2: Express \( \frac{dy}{dx} \).
\[ \frac{dy}{dx}=\frac{x+y}{x-y}. \]
This is a homogeneous differential equation because numerator and denominator are both homogeneous expressions of degree 1.

Step 3: Use substitution \( y=vx \).
Let:
\[ y=vx. \]
Then:
\[ \frac{dy}{dx}=v+x\frac{dv}{dx}. \]

Step 4: Substitute in the differential equation.
\[ v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}. \]
Taking \(x\) common in numerator and denominator:
\[ v+x\frac{dv}{dx}=\frac{1+v}{1-v}. \]

Step 5: Separate the variables.
\[ x\frac{dv}{dx}=\frac{1+v}{1-v}-v. \]
\[ x\frac{dv}{dx}=\frac{1+v-v+v^2}{1-v}. \]
\[ x\frac{dv}{dx}=\frac{1+v^2}{1-v}. \]
So,
\[ \frac{1-v}{1+v^2}dv=\frac{dx}{x}. \]

Step 6: Integrate both sides.
\[ \int \frac{1-v}{1+v^2}dv=\int \frac{dx}{x}. \]
\[ \int \frac{1}{1+v^2}dv-\int \frac{v}{1+v^2}dv=\log x+C. \]
\[ \tan^{-1}v-\frac{1}{2}\log(1+v^2)=\log x+C. \]

Step 7: Substitute \( v=\frac{y}{x} \) and simplify.
\[ \tan^{-1}\left(\frac{y}{x}\right)-\frac{1}{2}\log\left(1+\frac{y^2}{x^2}\right)=\log x+C. \]
\[ \tan^{-1}\left(\frac{y}{x}\right)=\log x+\frac{1}{2}\log\left(\frac{x^2+y^2}{x^2}\right)+C. \]
\[ \tan^{-1}\left(\frac{y}{x}\right)=\frac{1}{2}\log(x^2+y^2)+C. \]
Taking exponential on both sides:
\[ e^{\tan^{-1}\left(\frac{y}{x}\right)}=c\sqrt{x^2+y^2}. \]
Final Answer:
\[ \boxed{e^{\tan^{-1}\left(\frac{y}{x}\right)}=c\sqrt{x^2+y^2}}. \]