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Thread: How to solve this differential equation?

  1. #1
    Senior Member Vinod's Avatar
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    How to solve this differential equation?

    Hello,
    How to solve this differential equation $\frac{dy}{dx}=\frac{x+4y-3}{2x+3y-5}$ If we assume $u(x)=\frac{y(x)-\frac15}{x-\frac{11}{5}}$ how to proceed further?
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  2. #2
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    Re: How to solve this differential equation?

    if we let $\displaystyle v=x-\frac{11}{5}$ and $\displaystyle w=y-\frac{1}{5}$

    we get

    $\displaystyle \frac{\text{dw}}{\text{dx}}=\frac{\text{dy}}{\text {dx}}=\frac{x+4y-3}{2x+3y-5}=\frac{v+4 w}{2 v+3 w}=\frac{1+4u}{2+3u}$

    on the other hand $\displaystyle w=u v$ so

    $\displaystyle \frac{\text{dw}}{\text{dx}}=u +v \frac{\text{du}}{\text{dx}}=u+\left(x-\frac{11}{5}\right)\frac{\text{du}}{\text{dx}}$
    Thanks from Vinod and topsquark
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  3. #3
    Senior Member Vinod's Avatar
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    Re: How to solve this differential equation?

    Quote Originally Posted by Idea View Post
    if we let $\displaystyle v=x-\frac{11}{5}$ and $\displaystyle w=y-\frac{1}{5}$

    we get

    $\displaystyle \frac{\text{dw}}{\text{dx}}=\frac{\text{dy}}{\text {dx}}=\frac{x+4y-3}{2x+3y-5}=\frac{v+4 w}{2 v+3 w}=\frac{1+4u}{2+3u}$

    on the other hand $\displaystyle w=u v$ so

    $\displaystyle \frac{\text{dw}}{\text{dx}}=u +v \frac{\text{du}}{\text{dx}}=u+\left(x-\frac{11}{5}\right)\frac{\text{du}}{\text{dx}}$
    Hello,
    What would be the interval of validity for x and what would be the initial condition if we assume C=1.
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  4. #4
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    Re: How to solve this differential equation?

    We know that the initial value problem

    $\displaystyle y'(x)=\frac{x+4y-3}{2x+3y-5}$ , $\displaystyle y\left(x_0\right)=y_0$

    where $\displaystyle \left(x_0,y_0\right)$ is not on the line $\displaystyle 2x+3y-5=0$,

    has a unique solution in some interval containing $\displaystyle x_0$
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