Question

Reduce the equation

${\mathit{x}}^{\mathbf{2}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\mathbf{2}\mathit{x}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{0}$

to a differential equation with constant coefficients in $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{z}}^{\mathbf{2}}}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{z}}$, and y by the change of variable $\mathit{x}\mathbf{=}{\mathit{e}}^{\mathbf{x}}$. (See Chapter 8, Section 7d.)

Short answer

Hence, the required differential equation is $\frac{d^{2}y}{d{z}^2}+\frac{dy}{dz}-5y=0$.

Step-by-step solution

Higher order derivatives

If any function $f\left(x\right)$ has its derivative with respect to x that is $f\text{'}\left(x\right)$, then further differentiation of $f\text{'}\left(x\right)$ with respect to x results higher order derivative of $f\left(x\right)$.

Differentiate the given equation

The given equation is:

$x^2\frac{d^{2}y}{d{x}^2}+2x\frac{dy}{dx}-5y=0$

Now, we have$x=e^z\Rightarrow \frac{dx}{dz}=e^z\Rightarrow \frac{dz}{dx}=e^{-z}$. Then,

$$\begin{gathered} \frac{dy}{dx}=\frac{dy}{dz}\times \frac{dz}{dx} \\ =e^{-z}\frac{dy}{dz} \end{gathered}$$

And,

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(e^{-z}\frac{dy}{dz}\right)=\frac{d}{dz}\left(e^{-z}\frac{dy}{dz}\right)\frac{dz}{dx} \\ =\left(e^{-z}\frac{d^{2}y}{d{z}^2}-e^{-z}\frac{dy}{dz}\right)e^{-z} \\ =e^{-2z}\left(\frac{d^{2}y}{d{z}^2}-\frac{dy}{dz}\right) \\ {x}^2\frac{d^{2}y}{d{x}^2}=\frac{d^{2}y}{d{z}^2}-\frac{dy}{dz} \end{gathered}$$

Solve the obtained derivatives

Substitute all these values in $x^2\frac{d^{2}y}{d{x}^2}+2x\frac{dy}{dx}-5y=0$, and we get:

$$\begin{gathered} x^2\frac{d^{2}y}{d{x}^2}+2x\frac{dy}{dx}-5y=0 \\ \left(\frac{d^{2}y}{d{z}^2}-\frac{dy}{dz}\right)+2\frac{dy}{dx}-5y=0 \\ \frac{d^{2}y}{d{z}^2}+\frac{dy}{dz}-5y=0 \end{gathered}$$

Hence, the required differential equation is $\frac{d^{2}y}{d{z}^2}+\frac{dy}{dz}-5y=0$.