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  • If  is the solution of the differential equation <img alt="\mathrm{x \frac{d y}{d

If \mathrm{y=y(x)} is the solution of the differential equation \mathrm{x \frac{d y}{d x}+2 y=x e^{x}, y(1)=0} then the local maximum value of the function \mathrm{z(x)=x^{2} y(x)-e^{x}, x \in \mathbb{R}} is :

Option: 1

\mathrm{1-e}


Option: 2

0


Option: 3

\mathrm{\frac{1}{2}}


Option: 4

\mathrm{\frac{4}{e}-e}


Answers (1)

best_answer

\begin{aligned} &x \frac{d y}{d x}+2 y=x e^{x} \\ &\frac{d y}{d x}+\left(\frac{2}{x}\right) y=e^{x} \\ &I f=e^{\int \frac{2}{x} d x}=x^{2} \\ &\therefore\left(x^{2}\right)=\int x^{2} e^{x} d x \\ &y\left(x^{2}\right)=e^{x}\left(x^{2}-2 x+2\right)+c \end{aligned}

\text{at} x=1, y=0

0=e+c \Rightarrow c=-e

\therefore y(x)=\frac{e^{x}\left(x^{2}-2 x+2\right)-e}{x^{2}}

\begin{aligned} \quad z(x) &=e^{x}\left(x^{2}-2 x+2\right)-e-e^{x} \\ z(x) &=e^{x}\left(x^{2}-2 x+1\right)-e \\ z^{\prime}(x) &=2 e^{x}(x-1)+e^{x}(x-1)^{2} \\ &=e^{x}(x-1)[x+1] \end{aligned}

local maximum of x=-1

z(-1)=e^{-1}(4)-e=\frac{4}{e}-e

Posted by

SANGALDEEP SINGH

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