Question

Verify that ${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{x}$ is a solution of $\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{.}$Use reduction of order to find a second solution ${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$in the form of an infinite series. Conjecture an interval of definition for${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$

Short answer

Proved and $y_2\left(x\right)=x{E}_i\left(x\right)-e^x$

Step-by-step solution

Verify the solution of given equation

• A second order differential equation is a type of differential equation that contains a derivative of a function of order 2 and no other higher-order derivatives of the function.
• It contains terms such as y", $\frac{d^{2}y}{d{x}^2}$, y" (x), and so on that indicate the function's second order derivative.

Solving the second order differential equation

We have the second order differential equation

$xy\text{'}\text{'}-xy\text{'}+y=0······\left(1\right)$

with an alleged solution $y_1\left(x\right)=x$and we have to verify that it is a solution for this D.E as the following technique :

Differentiate this alleged solution with respect to $x$then we have

$y_1\text{'}=1······\left(a\right)$

And

$y_1\text{'}\text{'}=0······\left(b\right)$ (b)

After that, substitute with $y_1\left(x\right)=x$and equations and into equation (1), then we have

$$\begin{gathered} x\times \left(0\right)-x\times \left(1\right)+x=0 \\ 0-x+x=0 \\ 0=0 \end{gathered}$$

Then $y_1\left(x\right)=x$is a solution of the given differential equation.

Again we have the second order differential equation

$xy\text{'}\text{'}-xy\text{'}+y=0$

Which is

$y\text{'}\text{'}-y\text{'}+\frac{1}{x}y=0$

As the form

$y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=0$

with a first solution $y_1\left(x\right)=x$and we have to obtain the second solution using the formula of reduction of order as

$$\begin{gathered} y_2\left(x\right)=y_1\left(x\right)\int \frac{e^{-\int p\left(x\right)dx}}{{\left(y_1\left(x\right)\right)}^2}dx \\ =x\int \frac{e^{-\int -1dx}}{x^2}dx \\ =x\int \frac{e^{\int dx}}{x^2}dx \\ =x\int \frac{e^x}{x^2}dx \end{gathered}$$

We have to solve the integration $\int \frac{e^x}{x^2}dx$using integration by parts as the following:

If we set $u=e^x,$then we have $du=e^{x}dx$and if we set $dv=\frac{1}{x^2}dx,$then we have $v=-\frac{1}{x},$ then we obtain

$$\begin{gathered} \int \frac{e^x}{x^2}dx=\left[u\times v-\int vdu\right] \\ =\left[-\frac{e^x}{x}+\int \frac{e^x}{x}dx\right] \\ =E_i\left(x\right)-\frac{e^x}{x} \end{gathered}$$

Then we obtain

$y_2\left(x\right)=x{E}_i\left(x\right)-e^x$

Therefore this second solution conjectures an interval $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$ and${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{x}{\mathit{E}}_{\mathbf{i}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{-}{\mathit{e}}^{\mathbf{x}}$