Using Problems 29 and 31b, show that equation (6.24) is correct.

Several Terms on the Right-Hand Side: Principle of Superposition So far we have brushed over a question which may have occurred to you: What do we do if there are several terms on the right-hand side of the equation involving different exponentials?

In Problem 33 to 38 , solve the given differential equations by using the principle of superposition [see the solution of equation (6.29) . For example, in Problem 33 , solve three differential equations with right-hand sides equal to the three different brackets. Note that terms with the same exponential factor are kept together; thus a polynomial of any degree is kept together in one bracket.

${\mathbf{y}}^{\mathbf{"}}\mathbf{-}\mathbf{5}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{2}{\mathbf{e}}^{\mathbf{x}}\mathbf{+}\mathbf{6}\mathbf{x}\mathbf{-}\mathbf{5}$

The differential equation given in the question is $y^{"}-5y^{\text{'}}+6y=2e^x+6x-5$

A general solution to the nth order differential equation is one that incorporates a significant number of arbitrary constants. If one uses the variable approach to solve a first-order differential equation, one must insert an arbitrary constant as soon as integration is completed.

The given differential equation is

$$\begin{gathered} y^{"}-5y^{\text{'}}+6y=2e^x+6x-5 \\ \left(D^2+5D+6\right)y=2e^x+6x-5 \end{gathered}$$

To find the roots use auxiliary equation.

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathbf{m}\mathbf{+}\mathbf{6}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{m}\mathbf{-}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{0} \\ \mathbf{m}\left(m-3\right)\mathbf{-}\mathbf{2}\left(m-3\right)\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{m}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{3} \end{gathered}$$

C.F here is Complementary function

$C.\mathbf{F}\mathbf{=}{\mathbf{C}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{2}\mathbf{x}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{x}}$

Solve R.H.S as 2 part one with exponential and one with linear

$$\begin{gathered} {\left(P.I\right)}_1=\frac{1}{D^2-5D+6}6x-5 \\ =\frac{1}{6\left({\displaystyle \frac{D^2}{6}}-{\displaystyle \frac{5D}{6}}+1\right)}6x-5 \\ =\frac{1}{6}{\left(1+\frac{D^2-5D}{6}\right)}^{-1}6x-5 \\ =\frac{1}{6}\left(1-\frac{D^2-5D}{6}\right)6x-5 \end{gathered}$$

Solve the problem further

$$\begin{gathered} \frac{1}{6}\left(6x\right)\Rightarrow {\left(P.I\right)}_1 \\ =x \end{gathered}$$

So overall value of P.I is below

$$\begin{gathered} P.I={\left(P.I\right)}_1+{\left((P.I\right)}_2 \\ P.I=x+e^x \end{gathered}$$

so complete solution is

$C.S=C_1{e}^{2x}+C_2{e}^{3x}+x+e^x$

Hence the solution of the differential equation can be written as