[SOLVED] Differentiation

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Feb 28th 2009, 02:25 PM
ronaldo_07

[SOLVED] Differentiation

Find the general solution of the second-order inhomogeneous differential equation

y''+3y'+2y= $e^{2x}cos(x)$

So far I have:

$y(x)=Ae^{2x}sin(3x) +Be^{2x}cos(3x)$

$\frac{dy}{dx}= \frac{1}{5}(e^{2x}cos(x) +3e^{2x} sin(x))$

$\frac{d^{2}y}{dx^{2}}= \frac{1}{25}(-e^{2x}cos(x) +7e^{2x} sin(x))$

After this I substituted $\frac{dy}{dx} and \frac{d^{2}y}{dx^{2}}$ in to the origional equation. I'm not sure if what im doing is correct. And after this step I have no Idea what to do. Please Help, Thanks :)
Feb 28th 2009, 02:40 PM
Mush

Quote:

Originally Posted by ronaldo_07

Find the general solution of the second-order inhomogeneous differential equation

y''+3y'+2y= $e^{2x}cos(x)$

So far I have:

$y(x)=Ae^{2x}sin(3x) +Be^{2x}cos(3x)$

$\frac{dy}{dx}= \frac{1}{5}(e^{2x}cos(x) +3e^{2x} sin(x))$

$\frac{d^{2}y}{dx^{2}}= \frac{1}{25}(-e^{2x}cos(x) +7e^{2x} sin(x))$

After this I substituted $\frac{dy}{dx} and \frac{d^{2}y}{dx^{2}}$ in to the origional equation. I'm not sure if what im doing is correct. And after this step I have no Idea what to do. Please Help, Thanks :)

First of all use the Auxiliary equation to get the complimentary function:

$\lambda^2 + 3\lambda + 2 = 0$

$(\lambda + 2)(\lambda+1) = 0$

Hence $\lambda = -2 \text{ or } \lambda = -1$

So your CF is :

$y_c(x) = Ae^{-2x} + Be^{-x}$

Then consider the homogenous part of the equation which you have done.

But why did your constants A and B disappear when you differentiated them?

Also, you have to use the product rule:

$y(x)=Ce^{2x}sin(3x) +De^{2x}cos(3x)$

$\frac{dy}{dx}= \frac{d}{dx}\bigg(Ce^{2x}sin(3x) +De^{2x}cos(3x)\bigg)$

$= C\frac{d}{dx}\bigg(e^{2x}sin(3x)\bigg) +D\frac{d}{dx}\bigg(e^{2x}cos(3x)\bigg)$

$= C\bigg(e^{2x}\frac{d}{dx}\big(sin(3x)\big) + \sin(3x)\frac{d}{dx}\big(e^{2x}\big)\bigg) +D\bigg(e^{2x}\frac{d}{dx}\big(cos(3x)) + \cos(3x)\frac{d}{dx}\big(e^{2x}\big)\bigg)$
Feb 28th 2009, 03:21 PM
ronaldo_07

Quote:

Originally Posted by ronaldo_07

Find the general solution of the second-order inhomogeneous differential equation

y''+3y'+2y= $e^{2x}cos(x)$

So far I have:

$y(x)=Ae^{2x}sin(3x) +Be^{2x}cos(3x)$ typo should be x not 3x

$\frac{dy}{dx}= \frac{1}{5}(e^{2x}cos(x) +3e^{2x} sin(x))$

$\frac{d^{2}y}{dx^{2}}= \frac{1}{25}(-e^{2x}cos(x) +7e^{2x} sin(x))$

After this I substituted $\frac{dy}{dx} and \frac{d^{2}y}{dx^{2}}$ in to the origional equation. I'm not sure if what im doing is correct. And after this step I have no Idea what to do. Please Help, Thanks :)

typo, and i missed out the constants by mistake
Feb 28th 2009, 04:05 PM
ronaldo_07

After differentiating and substitution in to the origional equation what is the next step?