JEE Mains · Maths · STD 12 - 9. differential equations
If \(\frac{d y}{d x}=\frac{x y}{x^{2}+y^{2}} ; y(1)=1 ;\) then a value of \(x\) satisfying \(\mathrm{y}(\mathrm{x})=\mathrm{e}\) is
- A \(\sqrt{2} \mathrm{e}\)
- B \(\frac{e}{\sqrt{2}}\)
- C \(\frac{1}{2} \sqrt{3} \mathrm{e}\)
- D \(\sqrt{3} \mathrm{e}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{3} \mathrm{e}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=\frac{x y}{x^{2}+y^{2}}\) Let \(\mathrm{y}=\mathrm{vx}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}\)…
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