JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x), y>0\), be a solution curve of the differential equation \(\left(1+x^2\right) d y=y(x-y) d x\). If \(y(0)=1\) and \(y(2 \sqrt{2})=\beta\), then
- A \(e ^{3 \beta^{-1}}=e(3+2 \sqrt{2})\)
- B \(e ^{\beta^{-1}}= e ^{-2}(5+\sqrt{2})\)
- C \(e^{\beta^{-1}}=e^{-2}(3+2 \sqrt{2})\)
- D \(e ^{3 \beta^{-1}}= e (5+\sqrt{2})\)
Answer & Solution
Correct Answer
(A) \(e ^{3 \beta^{-1}}=e(3+2 \sqrt{2})\)
Step-by-step Solution
Detailed explanation
\(\left(1+x^2\right) d y=y(x-y) d x\) \(y(0)=1 \cdot y(2 \sqrt{2})=\beta\) \(\frac{d y}{d x}=\frac{y x-y^2}{1+x^2}\) \(\frac{d y}{d x}+y\left(\frac{-x}{1+x^2}\right)=\left(\frac{-1}{1+x^2}\right) y^2\)…
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