Page 320 Problem 17

page 320 problem 17

Corollary 6.2.2 on page 314 gives

$\mathcal{L}\{y^{(4)}\}=s^{4}\mathcal{L}\{y\}-s^{3}y(0)-s^{2}y^{\prime}(0)-sy^{\prime\prime}(0)-y^{\prime\prime\prime}(0)=s^{4}\mathcal{L}\{y\}-s^{2}-1$

$\mathcal{L}\{y^{\prime\prime\prime}\}=s^{3}\mathcal{L}\{y\}-s$

$\mathcal{L}\{y^{\prime\prime}\}=s^{2}\mathcal{L}\{y\}-1$

$\mathcal{L}\{y^{\prime}\}=s\mathcal{L}\{y\}$

Thus we have $\mathcal{L}\{y\}[s^{4}-4s^{3}+6s^{2}-4s+1]-s^{2}-1+4s-6=0$

Then we have $\mathcal{L}\{y\}(s-1)^{4}=s^{2}-4s+7$

So $\frac{s^{2}-4s+7}{(s-1)^{4}}=\frac{A}{(s-1)}+\frac{B}{(s-1)^{2}}+\frac{C}{(s-1)^{3}}+\frac{D}{(s-1)^{4}}$

Then we get $s^{2}-4s+7=A(s-1)^{3}+B(s-1)^{2}+C(s-1)+D$.

Immediately, because there is no cubed term, we have $A=0$.

The only square term is $Bs^{2}$and so $B=1$.

$s^{2}-4s+7=s^{2}-2s+1+Cs-C+D$

Then $-4=-2+C$ and so $C=-2$.

Then $7=1+2+D$. So $D=4$.

Great so we have $\frac{1}{(s-1)^{2}}+\frac{-2}{(s-1)^{3}}+\frac{4}{(s-1)^{4}}$.

At this point we should recognize that we are dealing with number
11 in the table (or a combination of number 3 and number 14).

$\mathcal{L}^{-1}\left(\frac{1}{(s-1)^{2}}+\frac{-2}{(s-1)^{3}}+\frac{4}{(s-1)^{4}}\right)=\mathcal{L}^{-1}\left(\frac{1}{1!}\frac{1!}{(s-1)^{2}}+\frac{-2}{2!}\frac{2!}{(s-1)^{3}}+\frac{4}{3!}\frac{3!}{(s-1)^{4}}\right)$

$=\mathcal{L}^{-1}\left(\frac{1}{(s-1)^{2}}\right)-\mathcal{L}^{-1}\left(\frac{2!}{(s-1)^{3}}\right)+\frac{2}{3}\mathcal{L}^{-1}\left(\frac{3!}{(s-1)^{4}}\right)=te^{t}-t^{2}e^{t}+\frac{2}{3}t^{3}e^{t}$

For the first equality, I just move the constants in the numerator
out of the way and multiple and divide by the constants I need to
be there. Then in the second equality, I simplify the constants in
front and use the linearity of the inverse laplace transform.