Question

(a) Verify that${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{x}}$ is a solution of$\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{10}\mathbf{)}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{10}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{.}$

(b) Use (5) to find a second solution $\mathbf{y}\mathbf{=}\left(x\right)\mathbf{.}$Use a CAS to carry out the required integration.

(c)Explain, using Corollary (A) of Theorem 4.1.2. whythe second solution can be written compactly as${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{10}}\frac{\mathbf{1}}{\mathbf{n}\mathbf{!}}{\mathit{x}}^{\mathbf{n}}$

Short answer

(a) Proved

(b)$y_2=-10!-10!x-\frac{10!}{2!}{x}^2-\frac{10!}{3!}{x}^3-\frac{10!}{4!}{x}^4-\frac{10!}{5!}{x}^5-\frac{10!}{6!}{x}^6-\frac{10!}{7!}{x}^7-\frac{10!}{8!}{x}^8-10x^9-x^{10}$

(c) Proved

Step-by-step solution

Verifying 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.

Step 2(a): Solving the second order differential equation

Let us suppose

$y\left(x\right)=e^x$

Therefore, we have

$$\begin{gathered} y\left(x\right)=e^x \\ y\text{'}\left(x\right)=e^x \\ y\text{'}\text{'}\left(x\right)=e^x \end{gathered}$$

Using above facts we have

$$\begin{gathered} xy\text{'}\text{'}-(x+10)y\text{'}+10y=x\left(e^x\right)-\left(x+10\right)·e^x+10e^x \\ =x{e}^x-x{e}^x-10x{e}^x+10e^x \\ xy\text{'}\text{'}-(x+10)y\text{'}+10y=0 \end{gathered}$$

Hence it is Verified that ${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{x}}$ is a solution of $\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{10}\mathbf{)}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{10}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{.}$

Step 3(b):Find the second solution

We have given as

$x{y}^{\text{'}\text{'}}-(x+10)y^{\text{'}}+10y=0$

The second solution, according to Eq (5) from the book is given by

$y_2=y_1\left(x\right)\int \frac{e^{-\int P\left(x\right)dx}}{y_1^2\left(x\right)}dx$

where$P\left(x\right)$ is given by the comparison of the above equation with the following

$y\text{'}\text{'}+P\left(x\right)y\text{'}+Q\left(x\right)y=0$

Therefore, we have

$$\begin{gathered} P\left(x\right)=-\frac{(x+10)}{x} \\ Q\left(x\right)=\frac{10}{x} \end{gathered}$$

Hence, $y_2$is given by

$$\begin{gathered} y_2=y_1\left(x\right)\int \frac{e^{-\int P\left(x\right)dx}}{y_1^2\left(x\right)}dx \\ {y}_2=e^x\int \frac{e^{\int \frac{(x+10)}{x}}dx}{{\left(e^x\right)}^2}dx \\ {y}_2=e^x\int \frac{e^{\int \left(1+\frac{10}{x}\right)}dx}{e^{2x}}dx \\ {y}_2=e^x\int \frac{e^{\int \left(1+\frac{10}{x}\right)}dx}{e^{2x}}dx \end{gathered}$$

$$\begin{gathered} y_2=e^x\int \frac{e^{(x+10\log (x\left)\right)}}{e^{2x}}dx \\ {y}_2=e^x\int \frac{e^{\left(x+\log \left(x^{10}\right)\right)}}{e^{2x}}dx \\ {y}_2=e^x\int \frac{e^x·e^{\log \left(x^{10}\right)}}{e^{2x}}dx \\ {y}_2=e^x\int \frac{e^x·x^{10}}{e^{2x}}dx \\ {y}_2=e^x\int e^{-x}·x^{10}dx \end{gathered}$$

Solving above with integration by parts 10 times, we get

$y_2=e^x\int e^{-x}·x^{10}dx$

$$\begin{gathered} y_2=e^x·\left[-3628,800-3628.800x-1814400x^2-604800x^3-151200x^4\right.\left.-30240x^5-5040x^6-720x^7-90x^8-10x^9-x^{10}\right]e^x \\ {y}_2=-3628,800-3628.800x-1814400x^2-604800x^3-151200x^4-30240x^5-5040x^6-720x^7-90x^8-10x^9-x^{10} \end{gathered}$$

Hence the second solution is

${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{10}\mathbf{!}\mathbf{-}\mathbf{10}\mathbf{!}\mathit{x}\mathbf{-}\frac{\mathbf{10}\mathbf{!}}{\mathbf{2}\mathbf{!}}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{10}\mathbf{!}}{\mathbf{3}\mathbf{!}}{\mathit{x}}^{\mathbf{3}}\mathbf{-}\frac{\mathbf{10}\mathbf{!}}{\mathbf{4}\mathbf{!}}{\mathit{x}}^{\mathbf{4}}\mathbf{-}\frac{\mathbf{10}\mathbf{!}}{\mathbf{5}\mathbf{!}}{\mathit{x}}^{\mathbf{5}}\mathbf{-}\frac{\mathbf{10}\mathbf{!}}{\mathbf{6}\mathbf{!}}{\mathit{x}}^{\mathbf{6}}\mathbf{-}\frac{\mathbf{10}\mathbf{!}}{\mathbf{7}\mathbf{!}}{\mathit{x}}^{\mathbf{7}}\mathbf{-}\frac{\mathbf{10}\mathbf{!}}{\mathbf{8}\mathbf{!}}{\mathit{x}}^{\mathbf{8}}\mathbf{-}\mathbf{10}{\mathit{x}}^{\mathbf{9}}\mathbf{-}{\mathit{x}}^{\mathbf{10}}$

Step 4(c): Explain using the Corollary (A) of the theorem 

The Corollary (A) of the Theorem 4.1.2 is given as,

"A constant multiple $y=c_1·y_1\left(x\right)$of a solution$y_1\left(x\right)$ of a homogeneous linear differential equation is also a solution."

We have

$$\begin{gathered} y_2=-10!-10!x-\frac{10!}{2!}{x}^2-\frac{10!}{3!}{x}^3-\frac{10!}{4!}{x}^4-\frac{10!}{5!}{x}^5-\frac{10!}{6!}{x}^6-\frac{10!}{7!}{x}^7-\frac{10!}{8!}{x}^8-10x^9-x^{10} \\ {y}_2=-\frac{10!}{0!},-\frac{10!}{1!}x-\frac{10!}{2!}{x}^2-\frac{10!}{3!}{x}^3-\frac{10!}{4!}{x}^4-\frac{10!}{5!}{x}^5-\frac{10!}{6!}{x}^6-\frac{10!}{7!}{x}^7-\frac{10!}{8!}{x}^8-\frac{10!}{9!}{x}^9-\frac{10!}{10!}{x}^{10} \\ {y}_2=-\sum _{n=0}^{10}\frac{10!}{n!}{x}^n \\ {y}_2=-10!·\left[\sum _{n=0}^{10}\frac{1}{n!}{x}^n\right] \end{gathered}$$

Therefore, by The Corollary (A) of the Theorem 4.1.2, the below solution is also a solution${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{10}}\frac{\mathbf{1}}{\mathbf{n}\mathbf{!}}{\mathit{x}}^{\mathbf{n}}$