Question:
The solution of the differential equation \(y^{2}-x^{2}\frac{dy}{dx}=xy\frac{dy}{dx}\) is ...
The solution of the differential equation \(y^{2}-x^{2}\frac{dy}{dx}=xy\frac{dy}{dx}\) is ...
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Write dy/dx as a function of y/x, substitute y equal to vx, and separate the variables.
Updated On: Nov 12, 2026
- \(\frac{y}{x}+\log |y|=c\)
- \(\frac{y}{x}+\log |x|=c\)
- \(\frac{y^{2}}{x}+\log |y|=c\)
- \(\frac{y}{x^{2}}+\log |y|=c\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
The equation is homogeneous, so put \(y=vx\) and separate the variables.
Step 2: Write dy/dx
\(y^{2}=\frac{dy}{dx}(x^{2}+xy)\Rightarrow\frac{dy}{dx}=\frac{y^{2}}{x(x+y)}=\frac{v^{2}}{1+v}\).
Step 3: Substitute y = vx
\(v+x\frac{dv}{dx}=\frac{v^{2}}{1+v}\Rightarrow x\frac{dv}{dx}=-\frac{v}{1+v}\).
So \(\frac{1+v}{v}dv=-\frac{dx}{x}\).
Step 4: Integrate
\(\log|v|+v=-\log|x|+c\Rightarrow\log\left|\frac{y}{x}\right|+\log|x|+\frac{y}{x}=c\).
This gives \(\frac{y}{x}+\log|y|=c\).
Final Answer:
The solution is y/x + log|y| = c. \[ \boxed{\text{(A) }\dfrac{y}{x}+\log |y|=c} \]
The equation is homogeneous, so put \(y=vx\) and separate the variables.
Step 2: Write dy/dx
\(y^{2}=\frac{dy}{dx}(x^{2}+xy)\Rightarrow\frac{dy}{dx}=\frac{y^{2}}{x(x+y)}=\frac{v^{2}}{1+v}\).
Step 3: Substitute y = vx
\(v+x\frac{dv}{dx}=\frac{v^{2}}{1+v}\Rightarrow x\frac{dv}{dx}=-\frac{v}{1+v}\).
So \(\frac{1+v}{v}dv=-\frac{dx}{x}\).
Step 4: Integrate
\(\log|v|+v=-\log|x|+c\Rightarrow\log\left|\frac{y}{x}\right|+\log|x|+\frac{y}{x}=c\).
This gives \(\frac{y}{x}+\log|y|=c\).
Final Answer:
The solution is y/x + log|y| = c. \[ \boxed{\text{(A) }\dfrac{y}{x}+\log |y|=c} \]
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