JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution curve of the differential equation
\(x\left(x^2+e^x\right) d y+\left(e^x(x-2) y-x^3\right) d x=0, x \gt 0\) passing through the point \((1,0)\). Then \(y(2)\) is equal to :
- A \(\frac{4}{4-\mathrm{e}^2}\)
- B \(\frac{2}{2+\mathrm{e}^2}\)
- C \(\frac{2}{2-\mathrm{e}^2}\)
- D \(\frac{4}{4+\mathrm{e}^2}\)
Answer & Solution
Correct Answer
(D) \(\frac{4}{4+\mathrm{e}^2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & x\left(x^2+e^x\right) d y+\left(e^x(x-2) y-x^3\right) d x=0 \\ & x\left(x^2+e^x\right) \frac{d y}{d x}+e^x(x-2) y=x^3 \\ & \frac{d y}{d x}+\frac{e^x(x-2)}{x\left(x^2+e^x\right)} y=\frac{x^2}{x^2+e^x} \\ & \text { I.F. }=e^{\int…
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