State the initial value problem y"-5y'+6y=f(t) subject to

We are given the initial value problem:
$y^{″}-5y^{\prime }+6y=f(t)$
with initial conditions $y(0)=y^{\prime }(0)=0$. We will solve this problem using Laplace transform method.
**Step 1:**
Taking the Laplace transform of both sides of the equation, and using the properties of Laplace transform, we get:
$s^{2}Y(s)-sy(0)-y^{\prime }(0)-5(sY(s)-y(0))+6Y(s)=F(s)$
Substituting $y(0)=y^{\prime }(0)=0$ and simplifying, we get:
$s^{2}Y(s)-5sY(s)+6Y(s)=F(s)$
**Step 2:**
Now, we will solve for $Y(s)$:
$s^{2}Y(s)-5sY(s)+6Y(s)=F(s)$
$(s^2-5s+6)Y(s)=F(s)$
$(s-2)(s-3)Y(s)=F(s)$
$Y(s)=\frac{F(s)}{(s-2)(s-3)}$
**Step 3:**
We will now take the inverse Laplace transform of $Y(s)$ to obtain the solution $y(t)$:
$Y(s)=\frac{F(s)}{(s-2)(s-3)}$
$Y(s)=\frac{A}{s-2}+\frac{B}{s-3}$ (partial fraction decomposition)
$Y(s)=\frac{A(s-3)+B(s-2)}{(s-2)(s-3)}$
Equating the coefficients of the terms on both sides, we get:
$A-B=1$
$-3A+2B=0$
Solving for $A$ and $B$, we get $A=-2$ and $B=-1$.
Therefore, $Y(s)=\frac{-2}{s-2}+\frac{-1}{s-3}$.
Taking the inverse Laplace transform, we get:
$y(t)=-2e^{2t}-e^{3t}$
So, the solution to the initial value problem $y^{″}-5y^{\prime }+6y=f(t)$, with initial conditions $y(0)=y^{\prime }(0)=0$, is $y(t)=-2e^{2t}-e^{3t}$.