Solve the differential equation:    \frac{\text d y}{\text d x} - \frac{2x}{1 + x^2}y = x^2 + 2

 

 

 

 
 
 
 
 

Answers (1)

\frac{\text d y}{\text d x} - \frac{2x}{1 + x^2}y = x^2 + 2    [Given]

    P = \frac{-2x}{1 + x^2}    and     Q = x^2 + 2

    \text{IF} = e^{\int Pdx} = e^{\int \frac{-2x}{1+x^2}dx}

    \text{IF} = \frac{1}{1+x^2}

\therefore The solution of given differential equation is 

        y(IF) = \int (IF\times Q)dx

\Rightarrow y\left (\frac{1}{1+x^2} \right )= \int \frac{x^2 + 2}{1 + x^2}dx

\Rightarrow y\left (\frac{1}{1+x^2} \right )= \int \left ( 1+\frac{1}{1 + x^2} \right )dx

\Rightarrow y\left (\frac{1}{1+x^2} \right )= x + \tan^{-1}x + c

\Rightarrow y= (1 + x^2)(x + \tan^{-1}x + c)

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