# Find the general solution

• May 15th 2009, 02:00 AM
matty888
Find the general solution
Find the general solution to the first order linear diff. equation

dy/dx + 3xy = -2x
• May 15th 2009, 02:26 AM
Soroban
Hello, matty888!

Quote:

Find the general solution of: .$\displaystyle \frac{dy}{dx} + 3xy \:=\: -2x$

Integrating factor: .$\displaystyle I \;=\;e^{\int3x\,dx} \;=\;e^{\frac{3}{2}x^2}$

Multiply: .$\displaystyle e^{\frac{3}{2}x^2}\frac{dy}{dx} + 3xye^{\frac{3}{2}x^2} \;=\;-2xe^{\frac{3}{2}x^2}$

We have: .$\displaystyle \frac{d}{dx}\left(e^{\frac{3}{2}x^2}y\right) \;=\;-2xe^{\frac{3}{2}x^2}$

Integrate: .$\displaystyle e^{\frac{3}{2}x^2}y \;=\;-2\int e^{\frac{3}{2}x^2}(x\,dx)$

On the right side:

. . Let: $\displaystyle u \:=\:\tfrac{3}{2}x^2 \quad\Rightarrow\quad du \:=\:3x\,dx \quad\Rightarrow\quad x\,dx \:=\:\tfrac{1}{3}du$

. . Substitute: .$\displaystyle -2\int e^u\left(\tfrac{1}{3}du\right) \;=\;-\tfrac{2}{3}\int e^u\,du \;=\;-\tfrac{2}{3}e^u + C$

. . Back-substitute: .$\displaystyle -\tfrac{2}{3}e^{\frac{3}{2}x^2} + C$

The equation becomes: .$\displaystyle e^{\frac{3}{2}x^2}y \;=\;-\tfrac{2}{3}e^{\frac{3}{2}x^2} + C$

Therefore: .$\displaystyle \boxed{y \;=\;-\frac{2}{3} + Ce^{-\frac{3}{2}x^2}}$

• May 15th 2009, 02:50 AM
matty888
Thanks Soroban
Hi Soroban,