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    Verify Solution of Differential equation

    Hello, i didn't know whether to post this in the calculus section but since its in my differential equations book I'll start here.

    The problem asks to verify the indicated function is a answer of the differential equation.Verify Solution of Differential equation-1.1-23.jpg

    My memory on the natural e is rusty. Any guidance? I feel like this should be easier than it looks.

    Thanks
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    Re: Verify Solution of Differential equation



    y = e^{-x^2}\left[\int_0^x e^{t^2} \, dt + c_1\right]

    product rule to find \frac{dy}{dx} ...

    \frac{dy}{dx} = e^{-x^2} \cdot e^{x^2} - 2xe^{-x^2}\left[\int_0^x e^{t^2} \, dt + c_1\right]

    \frac{dy}{dx} = 1 - 2xy

    (1 - 2xy) + 2xy = 1
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    Re: Verify Solution of Differential equation

    Do you have to solve the integral? i used mathway for guidance and it gave me zero. Heres a link

    Mathway: Evaluate the Integral
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    Re: Verify Solution of Differential equation

    Quote Originally Posted by skeeter View Post


    y = e^{-x^2}\left[\int_0^x e^{t^2} \, dt + c_1\right]

    product rule to find \frac{dy}{dx} ...

    \frac{dy}{dx} = e^{-x^2} \cdot e^{x^2} - 2xe^{-x^2}\left[\int_0^x e^{t^2} \, dt + c_1\right]

    \frac{dy}{dx} = 1 - 2xy

    (1 - 2xy) + 2xy = 1
    Since it's first order linear, you could solve the DE using the Integrating Factor.

    \displaystyle \frac{dy}{dx} + 2x\,y = 1, the integrating factor is \displaystyle e^{\int{2x\,dx}} = e^{x^2}, so multiplying both sides of the DE by the integrating factor gives

    \displaystyle \begin{align*}e^{x^2}\frac{dy}{dx} + 2x\,e^{x^2}y &= e^{x^2} \\ \frac{d}{dx}\left(e^{x^2}y\right) &= e^{x^2} \\ e^{x^2}y &= \int{e^{x^2}\,dx} + C_1 \\ y &= e^{-x^2}\int{e^{x^2}\,dx} + C_1e^{-x^2} \end{align*}
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