Finding the initial equation Having trouble with this problem.
Find the solution to the initial value problem 
$$\frac {dy}{dt} + y = t^2$$ 
Where $$y(0) = 0 $$
Can someone help me get started?
 A: It seems that the equation is $\frac{dy}{dt}+y=t^2$. You can apply the method variation of constants. First you have to solve the homogeneous equations.
$y'+y=0$
$y'=-y\quad | :y$
$\frac{1}{y} \ dy=-dt$
Integration
$ln(y)=-t+c$
Thus the homogeneous solution is $y_c=C\cdot e^{-t}$.
We have to find the particular solution. We have to start with
$y_p=C(t)\cdot e^{-t}$ 
Now you differentiate $y_p(t)$ w.r.t $t$ and you´ll get $y_p'(t)=C'(t)\cdot e^{-t}-C(t)\cdot e^{-t}$
Inserting the expression for $y_p$ and $y_p'(t)$ into the origin differential equation gives
$C'(t)\cdot e^{-t}\underbrace{-C(t)\cdot e^{-t}+C(t)\cdot e^{-t}}_{=0}=t^2$
This is equal to $C'(t)=t^2\cdot e^{t}$ To get $C(t)$ you can apply the method of integration by parts, in this case two times. You´ll get
$C(t)=e^t(t^2-2t+2)$
Therefore the solution is
$y=y_c+y_p=C\cdot e^{-t}+t^2-2t+2$
Now use $y(0)=0$ to evaluate the value of C. 
A: I don't know if $t_2$ is a constant, or if its supposed to be $t^2$, so I guess ill just let $t_2 = a$ as the procedure for attaining the solution is the same.
Multiplying both sides by $e^t$ yields
\begin{align}
e^t \frac{dy}{dt} + e^ty &= a e^t\\
\frac{d(e^t y)}{dt} &= ae^t\\
y = e^{-t} \int ae^t dt
\end{align}
If $a$ is in fact a constant then the integration is trivial, if its not and it is rather $t^2$ then you can use integration by parts to evaluate the integral.
A: Slightly different approach:
Solve the homogeneous equations $\;y'+y=0\;$ , whose characteristic polynomial is $\;r+1=0\iff r=-1\implies y_h=Ce^{-t}\;$ is the general solution for the homogeneous eq.
Now, since the right hand of the dif. eq. is $\;t^2e^0\;$ , a particular solutions is given by
$$y(t)=(At^2+Bt+C)\implies y'(t)=2At+B\implies$$
$$y'+y=At^2+(2A+B)t+(B+C)=t^2\stackrel{\text{comparing coeff.}}\implies A=1,\,B=-2,\,C=2\implies$$
a particular solution to this eq. is
$$y_p=t^2-2t+2$$
and thus the general solution for the original eq. is
$$y(t):=y_h(t)+y_p(t)=Ce^{-t}+t^2-2t+2$$
Now use $\;y(0)=0\;$ to find the constant: $\;0=\;C+2\implies C=-2$
