IVP

[jsbeckton]

I'm studying my old tests for finals and can't remember how to get this one started:

Solve the IVP: \(\displaystyle y' - 2xy = e^{x^2 } ,y(0) = 1\)

Can someone help me get started? Do I need to seperate the variables or use some other technique? Thanks a lot.

 

[soroban]

Hello, jsbeckton!

Solve: \(\displaystyle \;y'\,-\,2xy\:=\:e^{x^2},\;y(0)\,=\,1\)

This one requires an Integrating Factor: \(\displaystyle \;I\:=\:e^{\int(-2x)dx} \;=\;e^{-x^2}\)


Multiply through by \(\displaystyle I:\;\;e^{-x^2}\cdot y'\,-\,2xe^{-x^2}\cdot y\:=\:e^{-x^2}\cdot e^{x^2}\)

. . And we have: \(\displaystyle \:\frac{d}{dx}\left(e^{-x^2}\cdot y\right)\:=\:1\)


Integrate: \(\displaystyle \:e^{-x^2}\cdot y \:= \:x\,+\,C\)

. . and we have: \(\displaystyle \:y\:=\:e^{x^2}(x\,+\,C)\)


Since \(\displaystyle y(0)\,=\,1\), we have: \(\displaystyle \,e^0(0 + C)\:=\:1\;\;\Rightarrow\;\;C\,=\,1\)

. . Therefore: \(\displaystyle \:y\;=\;e^{x^2}(x + 1)\)

 

[jsbeckton]

Thanks, I forgot about the Integrating factor.