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Solve the differential equation: (1+y^2)...

Solve the differential equation: `(1+y^2)dx=(tan^(-1)y-x)dy`

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To solve the differential equation \((1+y^2)dx=(\tan^{-1}y-x)dy\), we will follow these steps: ### Step 1: Rearranging the Equation We start by rewriting the given equation in the form of \(\frac{dx}{dy}\): \[ (1+y^2)dx = (\tan^{-1}y - x)dy \] Dividing both sides by \(dy\) and \(1+y^2\), we get: ...
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Knowledge Check

  • Solution of the differential equation (1+y^(2))dx =(tan^(-1)y-x)dy is

    A
    `x e^(tan^(-1)y) =(1-tan^(-1)y)e^(tan^(-1)y)+c`
    B
    `x e^(tan^(-1)y)=(tan^(-1)y -1) e^(tan^(-1)y)+c`
    C
    `x=tan^(-1)y-1+c e^(-tan^(-1)y)`
    D
    None of these

  • The solution of the differential equation (1+y^(2)) dx = (tan^(-1) y - x) dy is

    A
    ` x = tan ^(-1)y -1+Ce^(-tan^(-1)y)`
    B
    ` y = tan^(-1)y+1 +Ce^(-tan^(-1)y)`
    C
    ` x = tan^(-1)y +Ce^(-tan^(-1))y`
    D
    None of the above

  • The solution of the differential equations (1+y^2)+(x-e^(tan^(-1)y))(dy)/(dx)=0 is

    A
    `2xe^(tan^(-1)y)=e^(2tan^(-1)y)+k`
    B
    `xe^(tan^(-1)y)=tan^(-1)y+k`
    C
    `xe^(2tan^(-1)y)=e^(tan^(-1)y)+k`
    D
    `(x-2)=ke^(-tan^(-1)y)`

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