Question

Use the substitution$\mathit{t}\mathbf{=}\mathit{x}\mathbf{-}{\mathit{x}}_{\mathbf{0}}$to solve the given differential equation${\left(x-4\right)}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\left(x-1\right)\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution for the given differential equation is$y=c_1(x-4{)}^3+c_2(x-4{)}^3\ln |x-4|$

Step-by-step solution

Definition of differential equation

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

Step: 2 Find y:

The given differential equation is${\left(x-4\right)}^{2}y\text{'}\text{'}-5\left(x-1\right)y\text{'}+9y=0$

We have to use substitution,

$t=x-x_0$

Let’s consider

$t=x-4$

$t^{2}y\text{'}\text{'}-5ty\text{'}+9y=0$

The substitution $y=t^m$

$$\begin{gathered} t^{2}y\text{'}\text{'}-5ty\text{'}+9y=0 \\ {t}^2\left[m(m-1)t^{m-2}\right]-5t\left[m{t}^{m-1}\right]+9t^m=0 \\ m(m-1)t^m-5m{t}^m+9t^m=0 \\ {t}^m\left[m^2-m-5m+9\right]=0 \\ {t}^m\left[m^2-6m+9\right]=0 \end{gathered}$$

So, $t=x-4$≠ 0

Using the factorized method,

$m^2-6m+9=0$

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

$=\frac{-(-6)\pm \sqrt{{(-6)}^2-4\left(1\right)\left(9\right)}}{2\left(1\right)}$

$$\begin{gathered} =\frac{6\pm \sqrt{36-36}}{2} \\ x=\frac{6\pm \sqrt{0}}{2} \end{gathered}$$

This becomes x=3.

Use case 2: Repeated real roots

Substitute the value of$m_1=m_2=3i$

$y=c_1{t}^3+c_2{t}^3\ln t$

Re-substituting $t=x-4$, the solution is$y=c_1(x-4{)}^3+c_2(x-4{)}^3\ln |x-4|$