Beryneingmk39
2022-03-22
Answered

Detemine whether the equation given is a solution to the corresponding differential equation.

$y={C}_{1}{e}^{-1.25t}+{c}_{2}{e}^{-5.5t}-16;$

$y{}^{\u2033}+\frac{27}{4{y}^{\prime}}+\frac{55}{8y}=-110$

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memantangti17

Answered 2022-03-23
Author has **13** answers

Given:

$y{}^{\u2033}+\frac{27}{4{y}^{\prime}}+\frac{55}{8y}=-110$

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

$ay{}^{\u2033}+b{y}^{\prime}+cy=g\left(x\right)$

The general solution to $a\left(x\right)y{}^{\u2033}+b\left(x\right){y}^{\prime}+c\left(x\right)y=g\left(x\right)$ can be written as

$y={y}_{h}+{y}_{p}$

$y}_{h$ is the solution to the homogeneous ODE $a\left(x\right)y{}^{\u2033}+b\left(x\right){y}^{\prime}+c\left(x\right)y=0$

$y}_{p$, the particular solution, is any function that satisfies the non−homogeneous equation

Find yh by solving the $y{}^{\u2033}+\frac{27}{4{y}^{\prime}}+\frac{55}{8y}=0$

For an equation $ay{}^{\u2033}+b{y}^{\prime}+cy=0$, assume a solution of the form $e}^{\gamma \cdot t$

Rewrite the equation with $y={e}^{\gamma \cdot t}$

$\left({e}^{\gamma \cdot t}\right){}^{\u2033}+\frac{27}{4}{\left({e}^{\gamma \cdot t}\right)}^{\prime}+\frac{55}{8}\cdot {e}^{\gamma \cdot t}=0$

Taking common out:

${e}^{\gamma \cdot t}({\gamma}^{2}+\frac{27\cdot \gamma}{4}+\frac{55}{8})=0$

${\gamma}^{2}+\frac{27\cdot \gamma}{4}+\frac{55}{8}=0$

$8\cdot {\gamma}^{2}+55\cdot \gamma +55=0$

For a quadratic equation of the form $a{x}^{2}+bx+c=0$ the solutions are

$x}_{1,2}=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a$

For $a=8,\text{}b=54,\text{}c=55$:

$\gamma}_{1,2}=\frac{-54\pm \sqrt{{54}^{2}-4\cdot 8\cdot 55}}{2\cdot 8$

$\gamma}_{1,2}=\frac{-54\pm \sqrt{2916-1760}}{16$

$\gamma}_{1,2}=\frac{-54\pm \sqrt{1156}}{16$

$\gamma}_{1,2}=\frac{-54\pm 34}{16$

$\gamma}_{1,2}=\frac{-54+34}{16},\text{}\frac{-54-34}{16$

The solution to the quadratic equation are:

$\gamma =-\frac{5}{4},\text{}\gamma =-\frac{11}{2}$

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 $y{}^{\u2033}+\frac{27}{4{y}^{\prime}}+\frac{55}{8}=-110$

${y}_{p}=-16$

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

Answered 2022-03-31
Author has **2495** answers

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