The differential equation is as follows-
$$y′′+4y′-5y=te^t$$ $$y(0)=1,y′(0)=0$$
$L[{y′′+4y′-5y}]=L{y′′}+4L{y′}-5L{y}=[s^2L{y}−sy(0)−y′(0)]+4[sL{y}−y(0)]-5L{y}=L{y}(s^2+4s-5)+s+4$
$L(te^t)= \frac{1}{(s-1)^2}$
$L(y)=\frac{1}{(s-1)^2} + s+4 $
After this I can't find the to perform partial fraction decomposition. Can you help me please? Thanks!!
Since $(s^2+4s-5)=(s-1)(s+5)$ we have: $$Y(s)=\frac {s+4}{(s-1)(s+5)}+\frac 1 {(s-1)^3(s+5)}$$ $$Y(s)=g(s)+f(s)$$ The first fraction is easy to decompose: $$g(s)=\frac {s+4}{(s-1)(s+5)}=\frac A {s-1}+\frac B {s+5}$$ $A=\frac 5 6$ and $B=\frac 16$ $$\implies y_h(t)=\mathcal {L^{-1}}\{g(s)\}=\frac 5 6e^t+\frac 16e^{-5t}$$ It's the solution to the homogeneous equation with the given initial conditions.
For the particular solution here is a little help: $$f(s)=\frac 1 {(s-1)^3(s+5)}$$ $$f(s)=\frac {A}{(s-1)^3}+\frac {B}{(s-1)^2}+\frac {C}{(s-1)}+\frac {D}{(s+5)}$$ We have $$A=\frac 16 \text { And } D=-\frac 1 {6^3}$$ Note that since $D,C$ are the only fraction with $s^3$ at the numerator so we can deduce that $C=-D=\frac 1 {6^3}$ For B you need to calculate...