Answer to Question #216983 in Differential Equations for Lwazi Boyce

Question

y"(t) + 4y'(t) + 4y(t) = 4e^-2t, y(0) = -1, y'(0) = 4

Answer 1

Characteristic equation corresponding to the given homogeneous differential equation

$y''(t)+4y'(t)+4y=0$ is $r^2+4r+4=0$

$\Rightarrow (r+2)^2=0\Rightarrow r=-2,-2$

General solution to the homogeneous differential equation is

$y_{h}=c_{1}e^{-2t}+c_2te^{-2t}$

Use Method of variation of parameters to find the particular solution

Particular solution to the given differential equation is

$y_{p}=Au+Bv,u=e^{-2t},v={te^{-2t}}$ where

$A=\int \frac{-vg(t)}{uv'-vu'}dt,B=\int \frac{ug(t)}{uv'-vu'}dt,g(t)=4e^{-2t}$

Find $uv'-vu'$

Now

$uv'-vu'=(e^{-2t})\left [ e^{-2t}-2te^{-2t} \right ]-(te^{-2t})(-2e^{-2t})=e^{-4t}-2te^{-4t}+2te^{-4t}=e^{-4t}$

$A=\int \frac{-e^{-2t}(4e^{-2t})}{e^{-4t}}dt=-4\int dt=-4t$

$B=\int \frac{te^{-2t}(4e^{-2t})}{e^{-4t}}dt=4\int(t) dt=\frac{t^2}{2}$

Using the above, we get

$y_{p}=(-4t)(e^{-2t})+(\frac{t^2}{2})({te^{-2t}})=-4te^{-2t}+\frac{t^3}{2}e^{-2t}$

General solution to the given differential equation is

$y=y_{h}+y_{p}=c_{1}e^{-2t}+c_{2}te^{-2t}-4te^{-2t}+\frac{t^3}{2}e^{-2t}$

Find the value of $c_{1}$ and $c_{2}$ using the given initial conditions $y(0)=-1,y'(0)=4$

$y=c_{1}e^{-2t}+c_{2}te^{-2t}-4te^{-2t}+\frac{t^3}{2}e^{-2t}$

$\Rightarrow y'(t)=-2c_{1}e^{-2t}+c_{2}(e^{-2t}-2te^{-2t})-4e^{-2t}+8te^{-4t}+\frac{3t^2}{2}e^{-2t}-t^3e^{-2t}$

$y(0)=-1\Rightarrow c_{1}+0=-1\Rightarrow c_{1}=-1$

$y'(0)=4\Rightarrow -2c_{1}+c_{2}(1-0)-4(1)+8(0)+0-0=4$

$\Rightarrow -c_{1}+c_{2}=0\Rightarrow c_{2}=-1$

Therefore, solution to the given differential equation is

$y=-e^{-2t}-te^{-2t}-4te^{-2t}+\frac{t^3}{2}e^{-2t}$