Question

Solution of the differential equation \(xdy-ydx-\sqrt{x^2+y^2}\,dx=0\) is

• A. \(y-\sqrt{x^2+y^2}=Cx^2\)
• B. \(y+\sqrt{x^2+y^2}=Cx^2\)
• C. \(x+\sqrt{x^2+y^2}=Cy^2\)
• D. \(x-\sqrt{x^2+y^2}=Cy^2\)

Hint:

When equation contains \(xdy-ydx\), try converting into \(\frac{dy}{dx}\) form and substitute \(y=vx\) for homogeneity.

Answer 1

The Correct Option is B

Step 1: Rewrite the given differential equation.
\[ xdy-ydx-\sqrt{x^2+y^2}\,dx=0 \]
Bring \(dx\) terms together:
\[ xdy = ydx + \sqrt{x^2+y^2}\,dx \]
\[ x\frac{dy}{dx} = y + \sqrt{x^2+y^2} \]
Step 2: Recognize homogeneous form.
\[ \frac{dy}{dx}=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2} \]
Let:
\[ \frac{y}{x}=v \Rightarrow y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \]
Step 3: Substitute.
\[ v+x\frac{dv}{dx}=v+\sqrt{1+v^2} \]
\[ x\frac{dv}{dx}=\sqrt{1+v^2} \]
Step 4: Separate variables.
\[ \frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x} \]
Step 5: Integrate both sides.
\[ \sinh^{-1}(v)=\ln|x|+C \]
\[ \ln\left|v+\sqrt{1+v^2}\right|=\ln|x|+C \]
Step 6: Remove logarithm.
\[ v+\sqrt{1+v^2}=Cx \]
Step 7: Substitute \(v=\frac{y}{x}\).
\[ \frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}=Cx \]
Multiply by \(x\):
\[ y+\sqrt{x^2+y^2}=Cx^2 \]
Final Answer:
\[ \boxed{y+\sqrt{x^2+y^2}=Cx^2} \]