JEE Mains · Maths · STD 12 - 9. differential equations
If \(y=y(x)\) is the solution of the differential equation \(x \frac{d y}{d x}+2 y=x e^{x}, y(1)=0\) then the local maximum value of the function \(z(x)=x^{2} y(x)-e^{x}\), \(x \in R\) is
- A \(1- e\)
- B \(0\)
- C \(\frac{1}{2}\)
- D \(\frac{4}{ e }- e\)
Answer & Solution
Correct Answer
(D) \(\frac{4}{ e }- e\)
Step-by-step Solution
Detailed explanation
\(x \frac{d y}{d x}+2 y=x e^{x}\) \(\frac{d y}{d x}+\frac{2 y}{x}=e^{x}\) \(\text { I.F. }=x^{2}\) \(y \cdot x^{2}=\int x^{2} e^{x} d x\) \(=\int e ^{x}\left( x ^{2}+2 x -2 x -2+2\right) dx\) \(yx^{2}=e^{x}\left(x^{2}-2 x+2\right)+c\) \(y (1)=0\) \(0= e (1+0)+ c\) \(c =- e\)…
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