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

Consider a differencial equation of the form

$$y^{″}+p(t)y^{\prime }+q(t)y=g(t)$$

Clerarly, this is not homogeneous, as $g(t)\ne 0$.

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 $A{e}^{2x}$. As $y^{″}$ and $y^{\prime }$ 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 $A{e}^{2x}$, as the particular integral cannot be equal to the general solution. In such cases, you need to take the substitution as $Ax{e}^{2x}$.

Example

Solve the differential equation

$$\frac{d^{2}y}{d{x}^2}+\frac{2dy}{dx}+2y=4+e^{-2x}$$
Given that
$$y(0)=4,y^{\prime }(0)=-1$$

Solution

Take $y=e^{mx}$. Then

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

Hence the general form of the equation becomes

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

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

$y\left(0\right)=4.$ Then $4=P+2+1\to P=1.$
$y^{\prime }\left(0\right)=-1$. Then $$-e^{-x}(P\cos x+Q\sin x)+e^{-x}(Q\cos x-P\sin x)-2e^{-2x}=-1=>Q-P-2=-1\to Q-P=1\to Q=2.$$

Hence the final equation is $$y=e^{-x}(\cos x+2\sin x)+2+e^{-2x}$$