Trouble finding a laplace transform solution

Question

So, I've got a problem that is $y''-5y'+6y=e^t, y(0)=1, y'(0)=1$.

I found the $$F(s) = \frac{s^2-5s+4}{((s-1)(s-2)(s-3)}$$

turning it into $$F(s) = \frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s-3} \; ,$$

I get $A=0, B=2, \text{ and } \; C=-1$.

And when I transform $F(s) -> f(t)$ I get $f(t) = 2e^{2t}-e^{3t}$, which doesn't solve the problem

Can someone please help me?

Answer 1

Using Laplace transform, we have $$\mathscr{L}\left\{y''(t)-5y'(t)+6y(t)\right\}=\mathscr{L}\left\{e^{t} \right\}$$ $$\implies \mathscr{L}\left\{y''(t)\}-5\mathscr{L}\left\{y'(t)\right\}+6\mathscr{L}\left\{y(t)\right\}\right\}=\mathscr{L}\left\{e^{t} \right\}$$ Since,

• $\mathscr{L}\{e^{t}\}=\frac{1}{s-1}$,

• $\mathscr{L}\{y(t)\}=Y(S)$,

• $\mathscr{L}\{y'(t)\}=SY(S)-y(0)$,

• $\mathscr{L}\{y''(t)\}=S^2Y(S)-Sy(0)-y'(0)$

Hence,

$$\mathscr{L}\left\{y''(t)\}-5\mathscr{L}\left\{y'(t)\right\}+6\mathscr{L}\left\{y(t)\right\}\right\}=\mathscr{L}\left\{e^{t} \right\}$$ $$\implies \left(S^{2}Y(S)-Sy(0)-y'(0)\right)-5\left(SY(S)-y(0)\right)+6Y(S)=\frac{1}{S-1}$$ $$\implies Y(S)(S^{2}-5S+6)=\frac{1}{S-1}+Sy(0)+y'(0)-5y(0)$$ Using,

• $y(0)=y'(0)=1$

Hence $$Y(S)(S^{2}-5S+6)=\frac{1}{S-1}+Sy(0)+y'(0)-5y(0)$$ $$\implies Y(S)(S^{2}-5S+6)=\frac{1}{S-1}+S+1-5$$ $$\implies \boxed{Y(S)=\frac{S^{2}-5S\color{red}{+5}}{(S-1)(S-2)(S-3)}}$$ $$\implies Y(S)=\frac{1/2}{S-1}+\frac{1}{S-2}+\frac{-1/2}{S-3}$$

Since,

• $\mathscr{L}^{-1}\left\{\frac{1}{S-a}\right\}=e^{at}$,
• $\mathscr{L}^{-1}\{Y(S)\}=y(t)$ where $Y(S):=\mathscr{L}\{y(t)\}$

Hence,

$$Y(S)=\frac{1/2}{S-1}+\frac{1}{S-2}+\frac{-1/2}{S-3}$$ $$\implies \mathscr{L}^{-1}\{Y(S)\}=\mathscr{L}^{-1}\left\{\frac{1/2}{S-1}+\frac{1}{S-2}+\frac{-1/2}{S-3}\right\}$$ $$y(t)=\frac{1}{2}\mathscr{L}^{-1}\left\{\frac{1}{S-1}\right\}+\mathscr{L}^{-1}\left\{\frac{1}{S-2}\right\}-\frac{1}{2}\left\{\frac{1}{S-3}\right\}$$ $$\implies \boxed{y(t)=\frac{1}{2}e^{t}+e^{2t}-\frac{1}{2}e^{3t}}$$ Notice,

• $y(0)=y'(0)=1$.
• $y''(t)-5y'(t)+6y(t)\equiv e^{t}$.

so done.

Remark:

In general for smooth functions over $[0,+\infty[$ and exponential order we have

• $\mathscr{L}\{y^{(n)}(t)\}=S^{n}Y(s)-S^{n-1}y(0)-S^{n-2}y'(0)-\cdots-y^{(n-1)}(0)$, where $Y(S)=\mathscr{L}\{y(t)\}$.