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  • Help me solve this - Differential equations - JEE Main-9

The solution of the differential equation x\frac{dy}{dx}+2y=x^{2}\left ( x\neq 0 \right )  with y\left ( 1 \right )=1, is :

  • Option 1)

     y=\frac{4}{5}x^{3}+\frac{1}{5x^{2}}       

  • Option 2)

    y=\frac{x^{3}}{5}+\frac{1}{5x^{2}}

  • Option 3)

     y=\frac{x^{2}}{4}+\frac{3}{4x^{2}}     

  • Option 4)

    y=\frac{3}{4}x^{2}+\frac{1}{4x^{2}}

 

Answers (1)

best_answer

x\frac{dy}{dx}+2y=x^{2}

\Rightarrow \frac{dy}{dx}+\frac{2}{x}y=x

This D.E in the form of Linear D.E.

IF=e^{\int \frac{2}{x}dx}=e^{2\ln x}=x^2

Solution,

y\cdot x^{2}=\int x^{2}\cdot x\; dx=\frac{x^{4}}{4}+C

given    \left ( y\left ( 1 \right )=1 \right )

 1=\frac{1}{4}+C\Rightarrow C=\frac{3}{4}

yx^{2}=\frac{x^{4}}{4}+\frac{3}{4}

y=\frac{x^{2}}{4}+\frac{3}{4x^{2}}


Option 1)

 y=\frac{4}{5}x^{3}+\frac{1}{5x^{2}}       

Option 2)

y=\frac{x^{3}}{5}+\frac{1}{5x^{2}}

Option 3)

 y=\frac{x^{2}}{4}+\frac{3}{4x^{2}}     

Option 4)

y=\frac{3}{4}x^{2}+\frac{1}{4x^{2}}

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