[SOLVED] Differentiation

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

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


So far I have:

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

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

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

After this I substituted $\displaystyle \frac{dy}{dx} and \frac{d^{2}y}{dx^{2}}$ 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 :)

Quote:

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Originally Posted by ronaldo_07

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

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


So far I have:

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

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

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

After this I substituted $\displaystyle \frac{dy}{dx} and \frac{d^{2}y}{dx^{2}}$ 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 :)

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

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

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

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

So your CF is :

$\displaystyle 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:

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

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

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

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

Quote:

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Originally Posted by ronaldo_07

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

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


So far I have:

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

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

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

After this I substituted $\displaystyle \frac{dy}{dx} and \frac{d^{2}y}{dx^{2}}$ 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 :)

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typo, and i missed out the constants by mistake

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