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Solve this problem - Letbe the solution of the differential equation,Ifthenis equal to : - Differential equations - JEE Main

Let y=y(x) be the solution of the differential equation, x\frac{dy}{dx}+y=x\log_{e}x,\: \: \left ( x> 1 \right ). If 2y(2)=\log_{e}4-1, then y(e) is equal to : 

  • Option 1)

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

     

     

     

  • Option 2)

    -\frac{e}{2}

  • Option 3)

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

  • Option 4)

    \frac{e}{4}

Answers (1)
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A admin

 

Linear Differential Equation -

\frac{dy}{dx}+Py= Q

- wherein

P, Q are functions of x alone.

 

 

Linear Differential Equation -

Multiply by e^{SPdx}  which is the Integrating factor

- wherein

P is the function of x alone

 

x\frac{dy}{dx}+y=x\log_{e}x,\: \: \left ( x> 1 \right ).

\frac{dy}{dx}+\frac{y}{x}=\log_{e}x

I.F. = e^{\int \frac{dx}{x}}=x

yx=\int x\log_{e}x\: \: dx

=>xy=(\log_{e}x)\times \frac{x^{2}}{2}-\frac{x^{2}}{4}+C

Putting x = 2 we get C = 0

y=\frac{x}{2}\log_{e}x-\frac{x}{4}

y(e)=\frac{e}{2}-\frac{e}{4}=\frac{e}{4}


Option 1)

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

 

 

 

Option 2)

-\frac{e}{2}

Option 3)

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

Option 4)

\frac{e}{4}

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