# Ordinary Differential Equations/Nonhomogeneous second order equations:Method of undetermined coefficients

Consider a differencial equation of the form

$y''+p(t)y'+q(t)y=g(t)$ Clerarly, this is not homogeneous, as $g(t)\neq 0$ .

So, to solve this, we first proceeed as normal, but assume that the equation is homogeneous; set $g(t)=0$ for now. Then the first part of the solution pans like

$y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}.$ .

Now we need to find the particular integral. To do this, make an appropriate substitution that relates to what $g(t)$ is. For instance, if $g(t)=e^{2x}$ , then take substitution $Ae^{2x}$ . As $y''$ and $y'$ are multiples of $y$ in this case, you'll simply get a linear equation in $A$ . Then just plug the value of $A$ in the equation.

Hence the solution is

y = general solution + particular integral.

There is one important caveat which you should be aware though. In the previous example for instance, if the general solution already had $e^{2x}$ , the substitution cannot be $Ae^{2x}$ , as the particular integral cannot be equal to the general solution. In such cases, you need to take the substitution as $Axe^{2x}$ .

## Example

Solve the differential equation

${\frac {d^{2}y}{dx^{2}}}+{\frac {2dy}{dx}}+2y=4+e^{-2x}$

Given that
$y(0)=4,y'(0)=-1$

### Solution

Take $y=e^{mx}$ . Then

$(d^{2}y)/(dx^{2})+2dy/dx+2y\rightarrow (m+1)^{2}+1=0\rightarrow m+1=\pm i\rightarrow m=-1\pm i$

Hence the general form of the equation becomes

$y=e^{-x}\left(P\cos {x}+Q\sin {x}\right)$

Now, the particular integral has to be found. To do so, we consider RHS: $4+{2e}^{-2x}$ . The substation then becomes $A+Be^{-2x}$ . Then $y^{\prime }=-2Be^{-2x}$  and $y^{\prime \prime }=4Be^{-2x}$ . Then the equation reduces to $4B-4B+2\left(A+Be^{-2x}\right)=4+2e^{-2x}\rightarrow 2A+2Be^{-2x}=4+2e^{-2x}$ . Hence $A=2,B=1$ . The equation is now

$y=e^{-x}\left(P\cos {x}+Q\sin {x}\right)+2+e^{-2x}$

$y\left(0\right)=4.$  Then $4=P+2+1\rightarrow P=1.$
$y^{\prime }\left(0\right)=-1$ . Then

$-e^{-x}\left(P\cos {x}+Q\sin {x}\right)+e^{-x}\left(Q\cos {x}-P\sin {x}\right)-2e^{-2x}=-1=>\ Q-P-2=-1\rightarrow Q-P=1\rightarrow Q=2.$

Hence the final equation is

$y=e^{-x}\left(\cos {x}+2\sin {x}\right)+2+e^{-2x}$