Answer to Question #154623 in Differential Equations for James

Question

Solve:y``-3y`+2y=e²t;y(0)=3;y-1(0)=5 by laplace transform

Answer 1

Let us solve $y''-3y'+2y=e^{2t},\ y(0)=3,\ y'(0)=5$ using laplace transform:

$y\to Y,\ \ y'\to pY-3,\ \ y''\to p^2Y-3p-5,\ \ e^{2t}\to \frac{1}{p-2}$

$p^2Y-3p-5-3pY+9+2Y=\frac{1}{p-2}$

$Y(p^2-3p+2)-3p+4=\frac{1}{p-2}$

$Y(p^2-3p+2)=\frac{1}{p-2}+3p-4=\frac{3p^2-10p+9}{p-2}$

$Y=\frac{3p^2-10p+9}{(p^2-3p+2)(p-2)}=\frac{3p^2-10p+9}{(p-1)(p-2)^2}$

$\frac{3p^2-10p+9}{(p-1)(p-2)^2}=\frac{A}{p-1}+\frac{B}{p-2}+\frac{C}{(p-2)^2}=\frac{A(p-2)^2+B(p-1)(p-2)+C(p-1)}{(p-1)(p-2)^2}$

$3p^2-10p+9=Ap^2-4Ap+4A+Bp^2-3Bp+2B+Cp-C$

$3p^2-10p+9=(A+B)p^2+(-4A-3B+C)p+4A+2B-C$

It follows that we have the following system:

$\begin{cases} A+B=3\\ -4A-3B+C=-10\\ 4A+2B-C=9 \end{cases}$

$\begin{cases} A=2\\ B=1\\ C=1 \end{cases}$

Therefore, $Y=\frac{2}{p-1}+\frac{1}{p-2}+\frac{1}{(p-2)^2}$

Since $\frac{1}{p-1}\to e^t,\ \ \frac{1}{p-2}\to e^{2t},\ \ \frac{1}{(p-2)^2}\to te^{2t},$ we conclude that the solution of the differential equation is the following:

$y(t)=2e^t+e^{2t}+te^{2t}.$