Solve differential equations containing $x''$ Solve the equation $$x'' + 3x' = 20e^{2t}$$ if $x(0) = 0$ and $x'(0) = 1$. 
I am pretty sure $x$ must contain $e^{2t}$, but other than that, I'm not really sure how to proceed from here. Any help would really be appreciated!
 A: Hints: Put $y=x'$. The homogeneous  equation becomes $y'+3y=0$ whose solution is $y=ae^{-3t}$. Integrate to get $x$. A particular solution of the type $ce^{2t}$ exists. Add the particular solution to the general solution of the homogeneous  equation and the apply the intial conditions. 
Answer: $e^{-3t}-3+2e^{2t}$. 
A: You could use Laplace transform to solve a Cauchy problem like this.
First apply the Laplace transform to the equation:
$$x'' + 3x' = 20e^{2t}\Rightarrow \mathcal{L}\{x''(t)+3x'(t)\}(s)=\mathcal{L}\{20e^{2t}\}(s)$$
You will get
$$s^2\mathcal{L}\{x(t)\}(s)-sx(0)-x'(0)+3s\mathcal{L}\{x(t)\}(s)-3x(0)=\frac{20}{s-2}$$
Since you know the values of $x(0)$ and $x'(0)$, you can solve the equation to $x(t)$:
$$s^2\mathcal{L}\{x(t)\}(s)-1+3s\mathcal{L}\{x(t)\}(s)=\frac{20}{s-2}\Rightarrow$$
$$\Rightarrow x(t)=\mathcal{L}^{-1} \left(\frac{22-s}{s(s+3)(s-2)}\right)$$
Now, decompose the fraction into simple ones and apply the Laplace inverse transform to those. You will get the result. To me, this is the best way to solve a linear equation with initial conditions (when the Laplace transform of the functions is known), since you don't have to use neither the Undetermined coefficients method nor the Variation of parameters.
A: $$x'' + 3x' = 20e^{2t}\implies (D^2 + 3D)x = 20e^{2t}\tag1$$where $~D=\dfrac{dx}{dt}~$.
Let $~x=Ae^{mt}~$be a solution of the equation $(1)~$.
Auxiliary equation is $$m^2+3m=0\implies m=0,-3$$
Solution of the Homogeneous equation of $(1)$ i.e., of $~(D^2 + 3D)x =0~$ is $$x=c_1~+c_2~e^{-3t}$$where $~c_1,~c_2~$are constants.
So the Complementary Function  (C.F.) of equation $(1)$ is $$c_1~+c_2~e^{-3t}$$where $~c_1,~c_2~$are constants.
Particular Integral (P.I.) 
$~~~=~\dfrac{1}{D^2 + 3D}~20e^{2t}$
$~~~=20~e^{2t}~\dfrac{1}{2^2 + 3\cdot2}$
$~~~=2~e^{2t}$
So the general solution of equation $(1)$ is $$x=c_1~+c_2~e^{-3t}+2~e^{2t}$$where $~c_1,~c_2~$are constants.
If $x(0) = 0$ and $x'(0) = 1~$, then $~c_1=-3~$and $~c_2=1~$.
So the p[articular solution is $$x=-3~+e^{-3t}+2~e^{2t}$$
A: Hint:
Try your assumption ! With $x=e^{2t}$, you have
$$4e^{2t}+3\cdot2e^{2t}\stackrel?=20 e^{2t},$$ which is correct to a constant factor. This gives you a particular solution.

Now if you add another exponential, let $x=e^{at}+?e^{2t}$,
$$(a^2+3a)e^{at}+10\,?e^{2t}=20e^{2t},$$
you have an identity when $a(a+3)=0.$
Now up to you to find the particular solution that fulfills the initial conditions.
A: For your work: The solution is given by $$x(t)=-\frac{1}{3} c_1 e^{-3 t}+c_2+2 e^{2 t}$$
