Q20 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

To derive the general solutions given by equations (17)- (20)  for the non-homogeneous equation (16), complete the following steps.

(a) Substitute $\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathbf{a}}_{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}$ and the Maclaurin series into equation (16) to obtain 

$(2a_2-a_0)\mathbf{+}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\left[(k+2)(k+1)a_{k+2}-(k+1)a_k\right]{\mathbf{x}}^{\mathbf{k}}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}}{\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{!}}{\mathbf{x}}^{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}}$

(b) Equate the coefficients of like powers on both sides of the equation in part (a) and thereby deduce the equations

${\mathbf{a}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{,}{\mathbf{a}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{6}}\mathbf{+}\frac{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{3}}\mathbf{,}{\mathbf{a}}_{\mathbf{4}}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{8}}\mathbf{,}{\mathbf{a}}_{\mathbf{5}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{40}}\mathbf{+}\frac{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{15}}\mathbf{,}{\mathbf{a}}_{\mathbf{6}}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{48}}\mathbf{,}{\mathbf{a}}_{\mathbf{7}}\mathbf{=}\frac{\mathbf{19}}{\mathbf{5040}}\mathbf{+}\frac{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{105}}$ 

(c) Show that the relations in part (b) yield the general solution to (16) given in equations (17)-(20).

Step-by-Step Solution

Answer

(a) After substituting $\mathrm{y}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$ and Maclaurin series, we get the solution $\left({\mathrm{a}}_2-{\mathrm{a}}_0\right)+\sum _{\mathrm{k}=1}^{\infty }\left[(\mathrm{k}+1)(\mathrm{k}+2){\mathrm{a}}_{\mathrm{k}+2}-(\mathrm{k}+1){\mathrm{a}}_{\mathrm{k}}\right]{\mathrm{x}}^{\mathrm{k}}=\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}-\cdots$.

(b) After equating the coefficients, the deduced equations are

$$\begin{gathered} 2{\mathrm{a}}_2-{\mathrm{a}}_0=0\to {\mathrm{a}}_2=\frac{{\mathrm{a}}_0}{2} \\ 6{\mathrm{a}}_3-2{\mathrm{a}}_1=1\to {\mathrm{a}}_3=\frac{1}{6}+\frac{{\mathrm{a}}_1}{3} \\ 12{\mathrm{a}}_4-3{\mathrm{a}}_2=0\to {\mathrm{a}}_4=\frac{{\mathrm{a}}_0}{8} \\ 20{\mathrm{a}}_5-4{\mathrm{a}}_3=-\frac{1}{6}\to {\mathrm{a}}_5=\frac{1}{40}+\frac{{\mathrm{a}}_1}{15} \\ 30{\mathrm{a}}_6-5{\mathrm{a}}_4=0\to {\mathrm{a}}_6=\frac{{\mathrm{a}}_0}{48} \\ 42{\mathrm{a}}_7-6{\mathrm{a}}_5=\frac{1}{120}\to {\mathrm{a}}_7=\frac{19}{5040}+\frac{{\mathrm{a}}_1}{105} \end{gathered}$$

(c) We showed that the results are similar to equations (17)- (20).

Step 1: Define power series expansion.

The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a coefficient recurrence relation. 

A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable. It is generally given by the formula,

$\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathbf{a}}_{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}$

Step 2: Find the relation.

The equation given in (16) is: 

$\mathrm{y}\text{'}\text{'}-\mathrm{xy}\text{'}-\mathrm{y}=\mathrm{sinx}$

Use the formula 

$\mathrm{y}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$

Take derivative

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\infty }{\mathrm{na}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1} \\ \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-2} \end{gathered}$$

The Maclaurin series is $\mathrm{sinx}=\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}-\cdots$

Substitute in the above equation.

$$\begin{gathered} \sum _{\mathrm{n}=2}^{\infty }\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-2}-\mathrm{x}·\sum _{\mathrm{n}=1}^{\infty }{\mathrm{na}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}-\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}=\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}-\cdots \\ \sum _{\mathrm{k}=0}^{\infty }(\mathrm{k}+1)(\mathrm{k}+2){\mathrm{a}}_{\mathrm{k}+2}{\mathrm{x}}^{\mathrm{k}}-\sum _{\mathrm{k}=1}^{\infty }{\mathrm{ka}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}-\sum _{\mathrm{k}=0}^{\infty }{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}=\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}-\cdots \\ {\mathrm{a}}_2+\sum _{\mathrm{k}=1}^{\infty }(\mathrm{k}+1)(\mathrm{k}+2){\mathrm{a}}_{\mathrm{k}+2}{\mathrm{x}}^{\mathrm{k}}-\sum _{\mathrm{k}=1}^{\infty }{\mathrm{ka}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}-{\mathrm{a}}_0-\sum _{\mathrm{k}=1}^{\infty }{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}=\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}-\cdots \\ \left({\mathrm{a}}_2-{\mathrm{a}}_0\right)+\sum _{\mathrm{k}=1}^{\infty }\left[(\mathrm{k}+1)(\mathrm{k}+2){\mathrm{a}}_{\mathrm{k}+2}-(\mathrm{k}+1){\mathrm{a}}_{\mathrm{k}}\right]{\mathrm{x}}^{\mathrm{k}}=\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}-\cdots \end{gathered}$$

Hence the relation is $\left({\mathrm{a}}_2-{\mathrm{a}}_0\right)+\sum _{\mathrm{k}=1}^{\infty }\left[(\mathrm{k}+1)(\mathrm{k}+2){\mathrm{a}}_{\mathrm{k}+2}-(\mathrm{k}+1){\mathrm{a}}_{\mathrm{k}}\right]{\mathrm{x}}^{\mathrm{k}}=\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}-\cdots$.

Step 3: Find the equations of coefficients.

Expand the expression.

$$\begin{gathered} \left(2{\mathrm{a}}_2-{\mathrm{a}}_0\right)+\left(6{\mathrm{a}}_3-2{\mathrm{a}}_1\right)\mathrm{x}+\left(12{\mathrm{a}}_4-3{\mathrm{a}}_3\right){\mathrm{x}}^2+ +\left(20{\mathrm{a}}_5-4{\mathrm{a}}_3\right){\mathrm{x}}^3+\left(30{\mathrm{a}}_6-5{\mathrm{a}}_4\right){\mathrm{x}}^4 \\ +\left(42{\mathrm{a}}_7-6{\mathrm{a}}_5\right){\mathrm{x}}^5+\left(56{\mathrm{a}}_8-6{\mathrm{a}}_6\right){\mathrm{x}}^6+\cdots =\mathrm{x}-\frac{{\mathrm{x}}^3}{6}+\frac{{\mathrm{x}}^5}{120}-\frac{{\mathrm{x}}^7}{5040}+\cdots \end{gathered}$$

Equate the coefficients of like powers on both sides of the equation.

$$\begin{gathered} 2{\mathrm{a}}_2-{\mathrm{a}}_0=0\to {\mathrm{a}}_2=\frac{{\mathrm{a}}_0}{2} \\ 6{\mathrm{a}}_3-2{\mathrm{a}}_1=1\to {\mathrm{a}}_3=\frac{1}{6}+\frac{{\mathrm{a}}_1}{3} \\ 12{\mathrm{a}}_4-3{\mathrm{a}}_2=0\to {\mathrm{a}}_4=\frac{{\mathrm{a}}_0}{8} \\ 20{\mathrm{a}}_5-4{\mathrm{a}}_3=-\frac{1}{6}\to {\mathrm{a}}_5=\frac{1}{40}+\frac{{\mathrm{a}}_1}{15} \\ 30{\mathrm{a}}_6-5{\mathrm{a}}_4=0\to {\mathrm{a}}_6=\frac{{\mathrm{a}}_0}{48} \\ 42{\mathrm{a}}_7-6{\mathrm{a}}_5=\frac{1}{120}\to {\mathrm{a}}_7=\frac{19}{5040}+\frac{{\mathrm{a}}_1}{105} \end{gathered}$$

Step 4: Find the expression after expansion.

Substitute the above coefficients in the equation:

$$\begin{gathered} \mathrm{y}\left(\mathrm{x}\right)={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+\frac{{\mathrm{a}}_0}{2}{\mathrm{x}}^2+\left(\frac{1}{6}+\frac{{\mathrm{a}}_1}{3}\right){\mathrm{x}}^3+\frac{{\mathrm{a}}_0}{8}{\mathrm{x}}^4+\left(\frac{1}{40}+\frac{{\mathrm{a}}_1}{15}\right){\mathrm{x}}^5+\frac{{\mathrm{a}}_0}{48}{\mathrm{x}}^6+\left(\frac{19}{5040}+\frac{{\mathrm{a}}_1}{105}\right){\mathrm{x}}^7+\cdots \\ \mathrm{y}\left(\mathrm{x}\right)={\mathrm{a}}_0\left(1+\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{8}+\frac{{\mathrm{x}}^6}{48}+\cdots \right)+{\mathrm{a}}_1\left(\mathrm{x}+\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^5}{15}+\frac{{\mathrm{x}}^7}{105}+\cdots \right)+\frac{{\mathrm{x}}^3}{6}+\frac{{\mathrm{x}}^5}{40}+\frac{19{\mathrm{x}}^7}{5040}+\cdots \end{gathered}$$

Where we can write:

$$\begin{gathered} {\mathrm{y}}_1=1+\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{8}+\frac{{\mathrm{x}}^6}{48}+\cdots \\ {\mathrm{y}}_2=\mathrm{x}+\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^5}{15}+\frac{{\mathrm{x}}^7}{105}+\cdots \\ {\mathrm{y}}_{\mathrm{p}}=\frac{{\mathrm{x}}^3}{6}+\frac{{\mathrm{x}}^5}{40}+\frac{19{\mathrm{x}}^7}{5040}+\cdots \end{gathered}$$

Hence, we showed that the results are similar to equations (17)-(20).