Question

write down the form of the general solution $\mathit{y}\mathbf{=}{\mathit{y}}_{\mathbf{c}}\mathbf{+}{\mathit{y}}_{\mathbf{p}}$v of the given differential equation in the two cases $\mathit{\omega }\mathbf{\ne }\mathit{\alpha }$and $\mathit{\omega }\mathbf{=}\mathit{\alpha }$Do not determine the coefficients in ${\mathit{y}}_{\mathbf{p}}$

$\mathit{y}\mathbf{"}\mathbf{-}{\mathit{\omega }}^{\mathbf{2}}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{\alpha }\mathbf{x}}$

Short answer

The general solution of the given differential equation is

$$\begin{gathered} y=c_1{e}^{\omega x}+c_2{e}^{-\omega x}+A{e}^{\alpha x} \\ y=c_1{e}^{\omega x}+c_2{e}^{-\omega x}+Ax{e}^{\alpha x} \end{gathered}$$

Step-by-step solution

Define particular solution

A Particular Solution is a differential equation solution that is derived from the General Solution by assigning specific values to the random constants. Depending on the query, the requirements for finding the values of the random constants can be submitted to us as an Initial-Value Problem or Boundary Conditions.

Now calculate the homogenous solution 

Here Second order DE

$y"-{\omega }^{2}y=e^{\alpha x}.....................\left(1\right)$

the corresponding homogeneous equation $y"+{\omega }^{2}y=0$by assuming that $y=e^{mx}$then differentiate with respect to x then we have

$$\begin{gathered} y\text{'}=m{e}^{mx} \\ y"=m^2{e}^{mx}.............................\left(2\right) \end{gathered}$$

substitute with $y=e^{mx}$and equation (2) into the homogeneous equation of equation (1),

then we have

$$\begin{gathered} m^2{e}^{mx}-{\omega }^2{e}^{mx}=0 \\ \left(m^2-{\omega }^2\right)e^{mx}=0 \end{gathered}$$

Since $e^{mx}$cannot be equal 0 , then we can have the auxiliary equation as

$$\begin{gathered} m^2-{\omega }^2=0 \\ \left(m-\omega \right)\left(m+\omega \right)=0 \end{gathered}$$

Then we have the roots as

$m_1=\omega and{m}_2=-\omega$

Then we can obtain the corresponding homogeneous solution as

$y_c=c_1{e}^{\omega x}+c_2{e}^{-\omega x}$

Now calculate the particular solution       

(a) For $\alpha \ne \omega$

we can have the particular solution as

$y_p=A{e}^{\alpha x}$

we can obtain the general solution of the given differential equation as

$$\begin{gathered} y=y_c+y_p \\ =c_1{e}^{\omega x}+c_2{e}^{\omega x}+A{e}^{\alpha x} \end{gathered}$$

(b) For$\alpha =\omega$

since we have repeated roots, then we can have the particular solution as

$$\begin{gathered} y_p=Ax{e}^{\alpha x} \\ =Ax{e}^{\omega x} \end{gathered}$$