Question

Use$\mathit{y}\mathbf{=}{\left(x-x_0\right)}^{\mathbf{m}}$to solve the given differential equation.${\left(x+3\right)}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\left(x+3\right)\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{14}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution for the given differential equation is$y=c_1(x+3{)}^2+c_2(x+3{)}^7$

Step-by-step solution

Define of differential equation:

An equation with one or more derivatives of a function dy/dx= f(x).

Find y:

The given differential equation is:

${\left(x+3\right)}^{2}y\text{'}\text{'}-8\left(x+3\right)y\text{'}+14y=0$

We have to use substitution;

$y={\left(x-x_0\right)}^m$

Let’s consider

$$\begin{gathered} y=(x+3{)}^m \\ \frac{dy}{dx}=m(x+3{)}^{m-1} \\ \frac{d^{2}y}{d{x}^2}=m(m-1)(x+3{)}^{m-2} \end{gathered}$$

Substituting the given differential equation

$$\begin{gathered} (x+3{)}^{2}y\text{'}\text{'}-8(x+3)y\text{'}+14y=0 \\ (x+3{)}^2\left[m(m-1){(x+3)}^{m-2}\right]-8(x+3)\left[m{(x+3)}^{m-1}\right]+14(x+3{)}^m=0 \\ m(m-1)(x+3{)}^m-8m(x+3{)}^m+14(x+3{)}^m=0 \\ (x+3{)}^m\left(m^2-m-8m+14\right)=0 \\ (x+3{)}^m\left(m^2-9m+14\right)=0 \end{gathered}$$

So, $(x+3{)}^m$≠ 0

Using factorize method,

$m^2-9m+14=0$

$=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-(-9)\pm \sqrt{{(-9)}^2-4\left(1\right)\left(14\right)}}{2\left(1\right)}$

$=\frac{9\pm \sqrt{81-56}}{2}$

$=\frac{9\pm \sqrt{25}}{2}$

$=\frac{9\pm 5}{2}$

This becomes m₁ = 2, m₂ = 7.

Use case 1: Distinct real roots

Substitute the values

The solution is$y=c_1(x+3{)}^2+c_2(x+3{)}^7$