JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(\left(x^2-3 y^2\right) d x+3 x y d y=0, y(1)=1\). Then \(6 y^2( e )\) is equal to \(......\)
- A \(3 e ^2\)
- B \(e ^2\)
- C \(2 e^2\)
- D \(\frac{3 e ^2}{2}\)
Answer & Solution
Correct Answer
(C) \(2 e^2\)
Step-by-step Solution
Detailed explanation
\(\left(x^2-3 y^2\right) d x+3 x y d y=0\) \(\frac{d y}{d x}=\frac{3 y^2-x^2}{3 x y} \Rightarrow \frac{d y}{d x}=\frac{y}{x}-\frac{1}{3} \frac{x}{y}\) Put \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)…
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