diff eqn using integrating factor

• Dec 5th 2006, 06:31 PM
ashura
diff eqn using integrating factor
Want to know if my answer is correct;

solve the diff eqn $\displaystyle \left (u^2 + 1 \right) \, \frac{dv}{du} + 4uv = 3u$

$\displaystyle \frac{dv}{du} + \frac{4uv}{u^2 + 1} = \frac{3u}{u^2 + 1}$

$\displaystyle Compare \, \, \frac{dy}{dx} + Py = Q$

$\displaystyle \therefore \, P = \frac{4u}{u^2 + 1} \, \, \, and \, \, \, Q = \frac{3u}{u^2 + 1}$

$\displaystyle Integrating \, \, Factor = e^{\int Pdu} = e^{\int \frac{4u}{u^2 + 1}} = e^{2 \int \frac{2u}{u^2 + 1}} = e^{2ln(u^2 + 1)} = e^{ln(u^2 + 1)^2}$

$\displaystyle = (u^2 + 1)^2$

$\displaystyle v \cdot Int \, Factor \, = \int Q \cdot Int \, Factor \, du$

$\displaystyle \therefore \, \, v(u^2 + 1)^2 = \int \frac{3u}{u^2 + 1} \cdot (u^2 + 1)^2 du$

$\displaystyle v(u^2 + 1)^2 = \int 3u(u^2 + 1) du$

$\displaystyle v(u^2 + 1)^2 = \int (3u^3 + 3u) du$

$\displaystyle = \frac{3u^4}{4} + \frac{3u^2}{2} + C$

$\displaystyle = \frac{3u^4 + 6u^2}{4} + C$

$\displaystyle = \frac{3u^2(u^2 + 2)}{4} + C$

$\displaystyle v = \frac{3u^2(u^2 + 2)}{4(u^2 +1 )^2} + \frac{C}{(u^2 + 1) ^2}$

$\displaystyle v = \frac{3u^2(u^2 + 2) + 4C}{4(u^2 +1 )^2} \, \,ans$
• Dec 5th 2006, 06:35 PM
ThePerfectHacker
Quote:

Originally Posted by ashura

$\displaystyle v = \frac{3u^2(u^2 + 2) + 4C}{4(u^2 +1 )^2} \, \,ans$

Good job!

Just a comment.
$\displaystyle 4C$ you can write $\displaystyle C$.