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  • Let be the solution of the differential equation <img alt="\mathrm{y(x+1) d x-x^2 d y

Let \mathrm{y=f(x)} be the solution of the differential equation \mathrm{y(x+1) d x-x^2 d y=0, y(1)=e}. Then \mathrm{\lim _{x \rightarrow 0^{+}} f(x)} is equal to


 

Option: 1

\frac{1}{e^2}


 


Option: 2

e^2


Option: 3

0


Option: 4

\frac{1}{e}


Answers (1)

best_answer

\mathrm{ y(x+1) d x-x^2 d y=0 \text {, } }            \mathrm{ y(1)=e}

\begin{aligned} & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}(\mathrm{x}+1)}{\mathrm{x}^2} \\ \end{aligned}

\begin{aligned} & \Rightarrow \int \frac{d y}{y}=\int \frac{(x+1) d x}{x^2} \\ \end{aligned}

\begin{aligned} & \ell \text { ny }=\ell \mathrm{nx}-\frac{1}{\mathrm{x}}+\mathrm{c} \\ \end{aligned}

\begin{aligned} & \because \mathrm{y}(1)=\mathrm{e} \\ \end{aligned}

\begin{aligned} & \therefore 1=0-1+\mathrm{C} \Rightarrow \mathrm{C}=2 \\ \end{aligned}

\begin{aligned} & \text { Now, } \ell \text { ny }=\ell \mathrm{nx}-\frac{1}{\mathrm{x}}+2 \\ & \end{aligned}

\begin{aligned} & \Rightarrow \ell \ln \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=2-\frac{1}{\mathrm{x}} \\ & \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\mathrm{e}^{2-\frac{1}{\mathrm{x}}} \\ & \Rightarrow \mathrm{y}=\mathrm{x}, \mathrm{e}^{2-\frac{1}{\mathrm{x}}} \end{aligned}

\mathrm{So, \lim _{x \rightarrow 0^{+}} y=\lim _{x \rightarrow 0^{+}} x e^{2-\frac{1}{x}}=0}

Posted by

Suraj Bhandari

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