MHT CET · Maths · Differential Equations
The general solution of the differential equation \(x^2+y^2-2 x y \frac{d y}{d x}=0\) is (where \(C\) is a constant of integration.)
- A \(2\left(x^2-y^2\right)+x=C\)
- B \(x^2+y^2=C x\)
- C \(x^2-y^2=C x\)
- D \(x^2+y^2=C y\)
Answer & Solution
Correct Answer
(C) \(x^2-y^2=C x\)
Step-by-step Solution
Detailed explanation
\(x^2+y^2-2 x y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{x^2+y^2}{2 x y} \Rightarrow v+x~ \cdot\) \( \frac{d v}{d x}=\frac{1+v^2}{2 v} \)
\( \text { [Let } y=n x] \)
\( \Rightarrow \int \frac{2 v d v}{1-v^2}=\int \frac{d x}{x} \)
\( \Rightarrow-\log \left|1-v^2\right|+\log |c|=\log |x| \)
\( \Rightarrow \log \left|\frac{c}{1-v^2}\right|=\log |x| \)
\( \Rightarrow \frac{c}{1-\frac{y^2}{x^2}}=x \)
\( \Rightarrow c x=x^2-y^2\)
\( \text { [Let } y=n x] \)
\( \Rightarrow \int \frac{2 v d v}{1-v^2}=\int \frac{d x}{x} \)
\( \Rightarrow-\log \left|1-v^2\right|+\log |c|=\log |x| \)
\( \Rightarrow \log \left|\frac{c}{1-v^2}\right|=\log |x| \)
\( \Rightarrow \frac{c}{1-\frac{y^2}{x^2}}=x \)
\( \Rightarrow c x=x^2-y^2\)
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