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.