# Non-homogenous equation (Undetermined coefficients?)

I have the initial value problem $4y''+4y'+4y = 1$; $y(0) = 3$, $y'(0) = 7$. I used undetermined coefficients to solve but got a very nasty answer. Is this the best method to use?

For the general solution I got $y(t)=e^{-x/2}(c_1cos(\frac {\sqrt{3}}{2}x)+c_2sin((\frac {\sqrt{3}}{2}x))$

The undetermined coefficient part was pretty easy -- basically just setting A=1/4 to get that the homogeneous equation plus 1/4.

Solving was trickier, I got $y(t)=e^{-x/2}(3cos(\frac {\sqrt{3}}{2}x)+\frac {17}{\sqrt3}sin((\frac {\sqrt{3}}{2}x))$

If something went wrong I would assume it was in solving for the coefficients.

• It's best if you post your answer. If it looks very mathematical, then use this page: google.co.in/… for tips on how to write the answer. – астон вілла олоф мэллбэрг Dec 8 '16 at 3:10
• @Moo Looks similar! Did you get the same general solution I did? Solving the initial value problem got very technical and may have been were we differed. – matt_Vera Dec 8 '16 at 3:45

I think another good way is via Laplace transformation. Let's denote by $L_f(s)$ the Laplace transform of $f$ in $s \in \mathbb{C}$. We have $y''(x)+y'(x)+y(x)=\frac{1}{4}$ therefore by applying the Laplace transform on both sides

\begin{align} L_{y''+y'+y}(s)&=L_\frac{1}{4} (s) \\ \Leftrightarrow L_{y''}(s)+L_{y'}(s)+L_y(s)&=\frac{1}{4} L_1(s)=\frac{1}{4s} \end{align}

where we used that $L_1(s)=\frac{1}{s}$. Additionally there are these useful properties of Laplace transforms: \begin{align} L_{y'}&=sL_y(s)-y(0)&&=sL_y(s)-3 \\ L_{y''}&=s^2L_y(s)-sy(0)-y'(0)&&=s^2L_y(s)-3s-7 \end{align}

So we get by plugging in:

\begin{align} s^2L_y(s)-3s-7+sL_y(s)-3+L_y(s)&=\frac{1}{4s} \\ \Leftrightarrow (s^2+s+1) L_y(s)&=\frac{1}{4s}+3s+10 \\ \Leftrightarrow L_y(s)&=\frac{12s^2+40s+1}{4s(s^2+s+1)}=\frac{1}{4s}+\frac{39}{4(s^2+s+1)}+\frac{11s}{4(s^2+s+1)} \end{align}

Now is the critical step that is not always so easy. We have to apply the inverse Laplace transform. Essentially we have to compute functions $f,g$ such that $L_f(s)=\frac{39}{4(s^2+s+1)}$ and $L_g(s)=\frac{11s}{4(s^2+s+1)}$.

Then by the theorem of Lerch you get $y(x)=\frac{1}{4}+f(x)+g(x)$ as the solution of your ODE. Computing $f,g$ explicitly you get $f(x)=\frac{67}{4 \sqrt{3}} e^{-x/2} \sin\left(\frac{\sqrt{3} x}{2}\right)$ and $g(x)=\frac{11}{4} e^{-x/2} \cos\left(\frac{\sqrt{3} x}{2}\right)$

You can also solve this with variation of parameters which is pretty standard but I think this is gonna be pretty nasty with computations. Computing the Wronskians etc. You can have a look at this link to get an idea how to do this in you case: https://en.wikipedia.org/wiki/Variation_of_parameters#General_second_order_equation

If you want to check: I got for the homogeneous part $$y_h(x)=Ae^{-x/2} \cos\left(\frac{x\sqrt{3}}{2}\right)+Be^{-x/2}\sin\left(\frac{x\sqrt{3}}{2}\right)$$ (where $A$ and $B$ are to be determined with the initial values) and the particular solution $$y_p(x)=\frac{1}{4}.$$ Now you get $y(x)=y_h(x)+y_p(x)$ as the solution to your ODE.

You need to add particular and general homogeneous solution before computing the coefficients from the eigenvalues. $$y(x)=\frac14+e^{-x/2}\left(A\cos(\frac {\sqrt{3}}{2}x)+B\sin(\frac {\sqrt{3}}{2}x)\right) \\ 3=y(0)=\frac14+A$$ which gives $A=\frac{11}4$. For the derivative one gets $$y'(x)=e^{-x/2}\left((\frac{\sqrt{3}}{2}B-\frac12A)\cos(\frac{\sqrt{3}}{2}x)-(\frac{\sqrt{3}}{2}A+\frac12B)\sin(\frac {\sqrt{3}}{2}x)\right)\\ 7=y'(0)=(\frac{\sqrt{3}}{2}B-\frac12A)$$ so that $\sqrt3 B=14+\frac{11}4=\frac{67}4$. Thus the numbers are even less nice.