Answer
$y=\frac{1}{5}(e^{3t}+4e^{-2t})$
Work Step by Step
The given initial value problem is
$\Rightarrow y''-y'-6y=0$
Initial values are
$y(0)=1,y'(0)=-1$
let the problem has a solution $y=\phi(t)$.
Take the Laplace transform of the differential equation.
$\Rightarrow \mathcal{L}\{y''\}-\mathcal{L}\{y'\}-6\mathcal{L}\{y\}=0$
Substitute $\mathcal{L}\{y''\}=s^2\mathcal{L}\{y\}-sy(0)-y'(0)$
and $\mathcal{L}\{y'\}=s\mathcal{L}\{y\}-y(0)$.
$\Rightarrow s^2\mathcal{L}\{y\}-sy(0)-y'(0)-[s\mathcal{L}\{y\}-y(0)]-6\mathcal{L}\{y\}=0$
Simplify.
$\Rightarrow s^2\mathcal{L}\{y\}-sy(0)-y'(0)-s\mathcal{L}\{y\}+y(0)-6\mathcal{L}\{y\}=0$
Use distributive property.
$\Rightarrow (s^2-s-6)\mathcal{L}\{y\}+(1-s)y(0)-y'(0)=0$
Substitute $\mathcal{L}\{y\}=Y(s)$ and initial values.
$\Rightarrow (s^2-s-6)Y(s)+(1-s)(1)-(-1)=0$
Simplify.
$\Rightarrow (s^2-s-6)Y(s)+1-s+1=0$
$\Rightarrow (s^2-s-6)Y(s)+2-s=0$
Isolate $Y(s)$.
$\Rightarrow Y(s)=\frac{s-2}{s^2-s-6}$
Factor denominator on the right hand side.
$\Rightarrow Y(s)=\frac{s-2}{s^2-3s+2s-6}$
$\Rightarrow Y(s)=\frac{s-2}{s(s-3)+2(s-3)}$
$\Rightarrow Y(s)=\frac{s-2}{(s-3)(s+2)}$ ......(1)
The partial fraction decomposition is
$\Rightarrow Y(s)=\frac{a}{(s-3)}+\frac{b}{(s+2)}$ ......(2)
$\Rightarrow Y(s)=\frac{a(s+2)+b(s-3)}{(s-3)(s+2)}$
Equate both numerators from equation (1) and (2).
$s-2=a(s+2)+b(s-3)$
Simplify.
$s-2=as+2a+bs-3b$
$1s-2=(a+b)s+2a-3b$
we can write.
$a+b=1$ ......(3)
and $2a-3b=-2$......(4)
Multiply equation (3) by $3$ and add to the equation (4).
$3a+3b+2a-3b=3-2$
Simplify.
$5a=1$
$a=\frac{1}{5}$
Substitute back into equation (3).
$\frac{1}{5}+b=1$
Isolate $b$.
$b=1-\frac{1}{5}$
$b=\frac{4}{5}$
Substitute back values of $a$ and $b$ into equation (2).
$\Rightarrow Y(s)=\frac{(\frac{1}{5})}{(s-3)}+\frac{(\frac{4}{5})}{(s+2)}$
$\Rightarrow Y(s)=\frac{1}{5(s-3)}+\frac{4}{5(s+2)}$
By using linearity of the Laplace transform.
$\Rightarrow y=\phi(t)=\frac{1}{5}e^{3t}+\frac{4}{5}e^{-2t}$
Use distributive property.
$\Rightarrow y=\frac{1}{5}(e^{3t}+4e^{-2t})$.
