# Detemine whether the equation given is a solution

Detemine whether the equation given is a solution to the corresponding differential equation.
$y={C}_{1}{e}^{-1.25t}+{c}_{2}{e}^{-5.5t}-16;$
$y{}^{″}+\frac{27}{4{y}^{\prime }}+\frac{55}{8y}=-110$
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memantangti17

Given:

A second order linear, non−homogeneous ODE has the form of

The general solution to  can be written as
$y={y}_{h}+{y}_{p}$
${y}_{h}$ is the solution to the homogeneous ODE
${y}_{p}$, the particular solution, is any function that satisfies the non−homogeneous equation
Find yh by solving the
For an equation , assume a solution of the form ${e}^{\gamma \cdot t}$
Rewrite the equation with $y={e}^{\gamma \cdot t}$

Taking common out:
${e}^{\gamma \cdot t}\left({\gamma }^{2}+\frac{27\cdot \gamma }{4}+\frac{55}{8}\right)=0$
${\gamma }^{2}+\frac{27\cdot \gamma }{4}+\frac{55}{8}=0$
$8\cdot {\gamma }^{2}+55\cdot \gamma +55=0$
For a quadratic equation of the form $a{x}^{2}+bx+c=0$ the solutions are
${x}_{1,2}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
For :
${\gamma }_{1,2}=\frac{-54±\sqrt{{54}^{2}-4\cdot 8\cdot 55}}{2\cdot 8}$
${\gamma }_{1,2}=\frac{-54±\sqrt{2916-1760}}{16}$
${\gamma }_{1,2}=\frac{-54±\sqrt{1156}}{16}$
${\gamma }_{1,2}=\frac{-54±34}{16}$

The solution to the quadratic equation are:

For two real roots ${\gamma }_{1}\ne q{\gamma }_{2}$, the general solution takes the form:
$y={c}_{1}{e}^{{\gamma }_{1}\cdot t}+{c}_{2}{e}^{{\gamma }_{2}\cdot t}$
${y}_{h}={c}_{1}{e}^{-\frac{3}{4t}}+{c}_{2}{e}^{-\frac{11}{2t}}$
${y}_{h}={c}_{1}{e}^{1.25t}+{c}_{2}{e}^{-5.5t}$
Now, find ${y}_{p}$ that satisfies
${y}_{p}=-16$
The general solution $y={y}_{h}+{y}_{p}$ is:
$y={c}_{1}{e}^{1.25t}+{c}_{2}{e}^{-5.5t}-16$
So, $y={c}_{1}{e}^{1.25t}+{c}_{2}{e}^{-5.5t}-16$ is a solution of differential equation
$y\text{'}\text{'}+\frac{27}{4y\text{'}}+\frac{55}{8y}=-110$

Jeffrey Jordon