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  • Confused! kindly explain, - Differential equations - JEE Main

. Let y = y(x) be the solution of the differential equation \frac{\mathrm{dy} }{\mathrm{d} x}+2y=f\left ( x \right ), where  f\left ( x \right )=\begin{Bmatrix} 1 \; \; \: ,&x\equiv \left [ 0,1 \right ] \\ 0\;\; \; , & otherwise \end{Bmatrix}

If y\left ( 0 \right )=0,\; then\; y\left ( \frac{3}{2} \right )\; is\; :

  • Option 1)

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

     

     

     

  • Option 2)

    \frac{1}{2e}

  • Option 3)

    \frac{e^{2}-1}{e^{3}}

  • Option 4)

    \frac{e^{2}-1}{2e^{3}}

 

Answers (2)

best_answer

As we learned 

 

Solution of differential equations -

In order to obtain the solution of differential equation, we integrate it as many times as the order of the differential equation

-

 

 Solving \frac{\mathrm{dy} }{\mathrm{d} x}+2y=f\left ( x \right )

we will get    y=\frac{1}{2}-\frac{1}{2}e^{-2x}

y\left ( 1 \right )=\frac{1-e^{-2}}{2}=\frac{e^{2}-1}{2e^{2}}

for   x\equiv \left ( 1,\infty \right )\; \; \; \; \; ;\; \; \; \; \; e^{2x}y=c_{2}

c_{2}=\frac{e^{2}-1}{2}\Rightarrow y=\frac{e^{2}-1}{2}e^{-2x}\cdot x\equiv \left ( 1,\infty \right )

so for x=\frac{3}{2},\; \; \; \; y=\frac{e^{2}-1}{2e^{3}}

 


Option 1)

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

 

 

 

Option 2)

\frac{1}{2e}

Option 3)

\frac{e^{2}-1}{e^{3}}

Option 4)

\frac{e^{2}-1}{2e^{3}}

Posted by

Himanshu

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