4𝑦 ′′(𝑡) + 24𝑦 ′ (𝑡) + 37𝑦(𝑡)

To solve the given second-order linear homogeneous differential equation, we'll use the method of finding the characteristic equation. Let's denote $y(t)$ as the unknown function and differentiate it with respect to $t$ to find the derivatives.
Given differential equation:
$4y^{″}(t)+24y^{\prime }(t)+37y(t)=0$
Differentiating $y(t)$ with respect to $t$:
$y^{\prime }(t)\to y^{″}(t)$
$y^{″}(t)\to {y^{″}}^{\prime }(t)$
Substituting the derivatives back into the differential equation, we get:
$4{y^{″}}^{\prime }(t)+24y^{\prime }(t)+37y(t)=0$
The characteristic equation is obtained by assuming a solution of the form $y(t)=e^{rt}$, where $r$ is a constant. Substituting this into the differential equation, we have:
$4(r^3{e}^{rt})+24(r{e}^{rt})+37(e^{rt})=0$
Factoring out $e^{rt}$, we get:
$(4r^3+24r+37)e^{rt}=0$
Since $e^{rt}$ is never zero for any value of $t$, we must have:
$4r^3+24r+37=0$
To solve this cubic equation, we can use numerical methods or factorization. By inspection, we find that $r=-1$ is a solution. Dividing the cubic equation by $(r+1)$, we get a quadratic equation:
$(4r^3+24r+37)/(r+1)=4r^2-4r+37$
Setting this quadratic equation equal to zero and solving it, we find that there are no real solutions. Therefore, $r=-1$ is the only real solution.
Now, the general solution of the differential equation is given by:
$y(t)=c_1{e}^{rt}+c_{2}t{e}^{rt}+c_3{t}^2{e}^{rt}$
where $c_1$, $c_2$, and $c_3$ are constants to be determined. Since we have two initial conditions, $y(\pi )=1$ and $y^{\prime }(\pi )=0$, we can substitute these values into the general solution to find the specific solution.
Substituting $t=\pi$ into the general solution, we have:
$1=c_1{e}^{-\pi }+c_2\pi e^{-\pi }+c_3{\pi }^2{e}^{-\pi }$
Substituting $t=\pi$ and differentiating the general solution with respect to $t$, we have:
$0=-c_1{e}^{-\pi }+c_2{e}^{-\pi }+2c_3\pi e^{-\pi }$
Now we have a system of two equations with three unknowns ($c_1$, $c_2$, and $c_3$). This means that there are infinitely many solutions that satisfy the given conditions.
Therefore, the solution to the given differential equation with the initial conditions $y(\pi )=1$ and $y^{\prime }(\pi )=0$ is expressed by the general solution:
$y(t)=c_1{e}^{-t}+c_{2}t{e}^{-t}+c_3{t}^2{e}^{-t}$
where $c_1$, $c_2$, and $c_3$ can take any values that satisfy the initial conditions.