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Math Help - differential equation

  1. #1
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    differential equation

    please find the attachment for help

    please help in solving.

    Solve the initial value problem:

    (y2 cosx-3x2 y-2x) dx + (2ysinx-x3​+ ln y)dy =0 ; y(0)=e
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  2. #2
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    Re: differential equation

    Quote Originally Posted by ararsil View Post
    please find the attachment for help

    please help in solving.

    Solve the initial value problem:

    (y2 cosx-3x2 y-2x) dx + (2ysinx-x3​+ ln y)dy =0 ; y(0)=e
    the form this is in makes it obvious that it is supposed to be an exact differential equation

    $du=M(x,y) dx + N(x,y) dy=0$

    $M(x,y)=\dfrac {du}{dx}=y^2 \cos(x)-3x^2 y-2x$

    $N(x,y)=\dfrac {du}{dy}=2y\sin(x)-x^3+\ln(y)$

    integrating M over x for example

    $u(x,y)=y^2\cos(x)-x^3 y-x^2+f(y)$

    where $f(y)$ is some function of $y$ we don't know yet that disappears when taking the x derivative

    You can do the same thing with $N(x,y)$ over $y$ and eventual derive a solution for $u(x,y)$

    Your solution is then in implicit form

    $u(x,y)=C$ where $C$ is a constant of integration. Maybe this can be solved for y in terms of x, maybe not.
    Thanks from ararsil
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  3. #3
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    Re: differential equation

    Quote Originally Posted by romsek View Post
    the form this is in makes it obvious that it is supposed to be an exact differential equation

    $du=M(x,y) dx + N(x,y) dy=0$

    $M(x,y)=\dfrac {du}{dx}=y^2 \cos(x)-3x^2 y-2x$

    $N(x,y)=\dfrac {du}{dy}=2y\sin(x)-x^3+\ln(y)$

    integrating M over x for example

    $u(x,y)=y^2\cos(x)-x^3 y-x^2+f(y)$
    Typo: you mean $u(x,y)= y^2sin(x)- x^3y- x^2+ f(y)$

    where $f(y)$ is some function of $y$ we don't know yet that disappears when taking the x derivative

    You can do the same thing with $N(x,y)$ over $y$ and eventual derive a solution for $u(x,y)$

    Your solution is then in implicit form

    $u(x,y)=C$ where $C$ is a constant of integration. Maybe this can be solved for y in terms of x, maybe not.
    Thanks from romsek
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