Question

Use the substitution$\mathit{y}\mathbf{=}{\left(x-x_0\right)}^{\mathbf{m}}$to solve the given differential equation.${\left(x-1\right)}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\left(x-1\right)\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution for the given differential equation is:$y=(x-1)\left[c_1\cos \left(2\ln \right(x-1\left)\right)+c_2\sin \left(2\ln \right(x-1\left)\right)\right.$

Step-by-step solution

Define 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

${\left(x-1\right)}^{2}y\text{'}\text{'}-\left(x-1\right)y\text{'}+5y=0$

We have to use the substitution;

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

Let’s consider

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

Substituting the given differential equation

$$\begin{gathered} (x-1{)}^{2}y\text{'}\text{'}-(x-1)y\text{'}+5y=0 \\ (x-1{)}^2\left[m(m-1){(x-1)}^{m-2}\right] \end{gathered}$$

$(x-1{)}^2\left[m(m-1){(x-1)}^{m-2}\right]$

$-(x-1)\left[m{(x-1)}^{m-1}\right]+5(x-1{)}^m=0$

$m(m-1)(x-1{)}^m-m(x-1{)}^m+5(x-1{)}^m=0$

$(x-1{)}^m\left(m^2-m-m+5\right)=0$

$(x-1{)}^m\left(m^2-2m+5\right)=0$

So, $(x-1{)}^m$≠ 0

Using the factorized method,

$$\begin{gathered} m^2-2m+5=0 \\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \end{gathered}$$

$x=\frac{-(-2)\pm \sqrt{{(-2)}^2-4\left(1\right)\left(5\right)}}{2\left(1\right)}$

$x=\frac{2\pm \sqrt{4-20}}{2}$

$x=\frac{2\pm \sqrt{-16}}{2}$

$x=\frac{2\pm 4i}{2}$

$x=\frac{2}{2}\pm \frac{4i}{2}$

$x=1\pm 2i$

This becomes;

$$\begin{gathered} x=1+2i \\ x=1-2i \end{gathered}$$

Use case: 3 conjugate complex roots

Substitute the value of $m_1$ and$m_2$ with the identification $\alpha =1$and$\beta =2$

The solution is$y=(x-1)\left[c_1\cos \left(2\ln \right(x-1\left)\right)+c_2\sin \left(2\ln \right(x-1\left)\right)\right.$