Question

Use the Laplace transform and inverse Laplace transform to solve in the initial value problem y''-3y'+2y=4t-6 y(0)=1, y'(0)=2

Answer 1

Denote by $Y(s)$ the Laplace transform of $y(t)$.

$y''-3y'+2y=4t-6$

Take the Laplace transform of both sides:

$(s^2Y(s)-sy(0)-y'(0))-3(sY(s)-y(0))+2Y(s)=\dfrac4{s^2}-\dfrac6s$

Solve for $Y(s)$:

$(s^2Y(s)-s-2)-3(sY(s)-1)+2Y(s)=\dfrac4{s^2}-\dfrac6s$

$(s^2-3s+2)Y(s)=\dfrac4{s^2}-\dfrac6s+s-1$

$Y(s)=\dfrac{4-6s-s^2+s^3}{s^2(s^2-3s+2)}$

$Y(s)=\dfrac{4-6s-s^2+s^3}{s^2(s-1)(s-2)}$

Decompose the right side into partial fractions: we're looking for constants $a_1,\ldots,a_4$ such that

$\dfrac{4-6s-s^2+s^3}{s^2(s-1)(s-2)}=\dfrac{a_1}s+\dfrac{a_2}{s^2}+\dfrac{a_3}{s-1}+\dfrac{a_4}{s-2}$

$4-6s-s^2+s^3=a_1s(s-1)(s-2)+a_2(s-1)(s-2)+a_3s^2(s-2)+a_4s^2(s-1)$

Expanding on the right side gives

$2a_2+(2a_1-3a_2)s-(3a_1-a_2+2a_3+a_4)s^2+(a_1+a_3+a_4)s^3$

and matching up coefficients gives the system

$\begin{cases}2a_2=4\\2a_1-3a_2=-6\\3a_1-a_2+2a_3+a_4=1\\a_1+a_3+a_4=1\end{cases}\implies a_1=0,a_2=2,a_3=2,a_4=-1$

So we have

$Y(s)=\dfrac2{s^2}+\dfrac2{s-1}-\dfrac1{s-2}$

and taking the inverse transform of both sides is trivial, giving

$\boxed{y(t)=2t+2e^t-e^{2t}}$