# Thread: Verify Solution of Differential equation

1. ## 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.

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

Thanks

2. ## 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 $\frac{dy}{dx}$ ...

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

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

$(1 - 2xy) + 2xy = 1$

3. ## 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

4. ## Re: Verify Solution of Differential equation

Originally Posted by skeeter

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

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*}