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  • Let be the solution of the differential equation  <img alt="x \log _e x \frac{d y}{d x}+y=x^2 \log _e x,

Let y=y(x) be the solution of the differential equation  x \log _e x \frac{d y}{d x}+y=x^2 \log _e x,(x>1). If y(2)=2, then y(e)  is equal to

Option: 1

\frac{1+e^2}{2}


Option: 2

\frac{4+e^2}{4}


Option: 3

\frac{2+e^2}{2}


Option: 4

\frac{1+e^2}{4}


Answers (1)

best_answer

D.E. \, \, \frac{d y}{d x}+\frac{1}{x \log x} y=x\\ Linear \, \, diff. \, \, eq { }^{\mathrm{n}} \, \, \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{py}=\mathrm{Q}

$$ \begin{aligned} & \text { If }=\mathrm{e}^{\int \mathrm{Pdx}} \\ &=\mathrm{e}^{\int \frac{1}{\mathrm{x} \log \mathrm{x}} \mathrm{dx}} \quad \begin{array}{l} \log \mathrm{x}=\mathrm{t} \\ \frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt} \end{array} \\ &=\mathrm{e}^{\int \frac{1}{\mathrm{t} \mathrm{dt}}}=\mathrm{e}^{\ell \mathrm{nt}}=\mathrm{t} \\ & \text { I.F. }=\ell \mathrm{nx} \end{aligned}

Solution of DE.

\begin{aligned} & \text { y. If }=\int q(\text { If }) d x+c \\ & y . \ell \mathrm{nx}=\int x .(\ell \operatorname{nx}) \mathrm{dx}+\mathrm{c} \\ & =\ell \mathrm{nx} \cdot \frac{\mathrm{x}^2}{2} \ell \mathrm{nx}-\frac{\mathrm{x}^2}{4}+\mathrm{C} \\ & \text { At } x=2, y=2 \\ & 2 \ell \mathrm{n} 2=\frac{4}{2} \ell \mathrm{n} 2-\frac{4}{4}+\mathrm{C} \Rightarrow \mathrm{C}=1 \\ & \therefore \quad \mathrm{y} \ell \mathrm{nx}=\frac{\mathrm{x}^2}{2} \ell \mathrm{nx}-\frac{\mathrm{x}^2}{4}+1 \\ & \end{aligned}

At  x=e

\begin{aligned} & y(e) \ell n e=\frac{e^2}{2} \ell \mathrm{n} e-\frac{e^2}{4}+1 \\ & y(e)=\frac{e^2}{4}+1 \end{aligned}

Posted by

Divya Prakash Singh

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