Solve the differential equation y double prime plus 2y prime plus y equals e to the 3x.

Step-by-step solution

1. 1

   Split the task in two. For a linear equation with constant coefficients the general solution is $y=y_h+y_p$: the general solution of the homogeneous equation plus any one particular solution of the full equation. Each half is found by a different technique.

2. 2

   Solve the homogeneous part. Set $y''+2y'+y=0$ and substitute $y=e^{rx}$:

   $r^2+2r+1=(r+1)^2=0\quad\Rightarrow\quad r=-1\ \text{(twice)}$

   A repeated root supplies only one exponential, so the second independent solution is that exponential times $x$:

   $y_h=(C_1+C_2x)e^{-x}$

3. 3

   Choose the right trial function. The right-hand side is $e^{3x}$. Because $3$ is not a root of $r^2+2r+1$, no resonance occurs and the plain guess $y_p=Ae^{3x}$ suffices. (Had the right side been $e^{-x}$, matching the repeated root, the trial would have needed an $x^2$ factor.)

4. 4

   Substitute and solve for $A$. With $y_p'=3Ae^{3x}$ and $y_p''=9Ae^{3x}$,

   $(9A+6A+A)e^{3x}=e^{3x}\quad\Rightarrow\quad 16A=1\quad\Rightarrow\quad A=\frac{1}{16}$

   The $16$ is no accident: it is the characteristic polynomial evaluated at the exponent, $p(3)=3^2+2(3)+1=16$.

5. 5

   Assemble the general solution.

   $y=(C_1+C_2x)e^{-x}+\frac{1}{16}e^{3x}$

6. 6

   Verify both pieces. For $y_p=\frac{1}{16}e^{3x}$: $\frac{9}{16}+\frac{6}{16}+\frac{1}{16}=\frac{16}{16}=1$, so it reproduces $e^{3x}$ ✓. For $y_h=xe^{-x}$: $y_h'=(1-x)e^{-x}$, $y_h''=(x-2)e^{-x}$, and $(x-2)+2(1-x)+x=0$, so it solves the homogeneous equation ✓.