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(x^2+x y)dy=(x^2+y^2)dx...

`(x^2+x y)dy=(x^2+y^2)dx`

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y= sin ^(-1)x show that (1-x^2) (d^2 y)/(dx^2) -x (dy)/(dx) =0

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Knowledge Check

  • Solution of y dx - x dy = x^2 ydx is :

    A
    `ye^(x^2)=cx^2`
    B
    `ye^(-x^2)=cx^2`
    C
    `y^2e^(x^2)=cx^2`
    D
    `y^2e^(-x^2)=cx^2`

  • The solution of x dx + y dy = x^(2)y dy - xy^(2) dx is

    A
    `x^(2)-1 = c(1+y^(2))`
    B
    `x^(2)+1 = c(1-y^(2))`
    C
    `x^(3) -1 = c (1 + y^(3))`
    D
    `x^(3)+1 = c (1-y^(3))`

  • If x+ y = tan^(-1) y and (d^(2)y)/(dx^(2)) =f (y) (dy)/(dx) , then f(y) =

    A
    `(-2)/(y^(3))`
    B
    `(2)/(y^(3))`
    C
    `(1)/(y)`
    D
    `(-1)/(y)`

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    If y= ( sin^(-1) x)^(2) , show that (1-x^(2) ) (d^2 y)/( dx^2) - x (dy)/(dx) =2

    The differential equations of all circle touching the x-axis at orgin is (a) (y^(2)-x^(2))=2xy((dy)/(dx)) (b) (x^(2)-y^(2))(dy)/(dx)=2xy ( c ) (x^(2)-y^(2))=2xy((dy)/(dx)) (d) None of these

    show that the given differential equation is homogeneous and solve it. x dy - y dx = sqrt(x^(2) + y^(2)) dx

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    If cos^(-1)((x^(2)-y^(2))/(x^(2)+y^(2))) = log a , then dy/dx =

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