Question

In Problems 35–64 solve the given differential equation by undetermined coefficients.

$\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{2}\mathbf{y}\text{'}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{4}{\mathbf{e}}^{\mathbf{x}}\mathbf{-}\mathbf{9}$

Short answer

$y=C_1{e}^{-x}+C_2{e}^{3x}-e^x+3$

Step-by-step solution

Definition of homogeneous differential equation

A homogeneous differential equation is an equation containing a differentiation and a function, with a set of variables.

Given data

Consider the differential equation as follows:

$y^{\text{'}\text{'}}-2y^{\text{'}}-3y=4e^x-9$

The objective is to solve the above differential equation by undetermined coefficients.

The solution of the corresponding homogeneous equation is as follows:

$\begin{array}{rcl}{m}^2-2m-3& =& 0\\ m^2-3m+m-3& =& 0\\ m(m-3)+1(m-3)& =& 0\\ (m-3)(m+1)& =& 0\\ m_1& =& -1\text{and}{m}_2=3\end{array}$

Thus, the complementary solution is, .$y_c=C_1{e}^{-x}+C_2{e}^{3x}$

Apply differential operator

From the properties of derivatives,

$\begin{array}{rcl}D\left(9\right)& =& 0\\ & & (D-1)\left(e^x\right)\text{=}0\end{array}$

Hence,$D(D-1)\left(4e^x-9\right)=0$

Apply$D(D-1)$ to the both sides of the given differential equation,

$\begin{array}{rcl}D(D-1)\left(D^2-2D-3\right)y& =& D(D-1)\left(4e^x-9\right)\\ D(D-1)\left(D^2-2D-3\right)y& =& 0\end{array}$

The auxiliary equation of the above equation is,

$\begin{array}{rcl}m(m-1)\left(m^2-2m-3\right)& =& 0\\ & \Rightarrow & m_1=-1,m_2=3,m_3=0,m_4=1\end{array}$

Thus, the general solution must be, $y=\underset{y_c}{\underbrace{C_1{e}^{-x}+C_2{e}^{3x}+C_3}}+C_4{e}^x$.

Solve for complementary solution

The first two terms of y represents the complementary solution, the last two terms denote the basic form of particular solution.

Thus,$y_p=C_3+C_4{e}^x$

$$\begin{gathered} y_p\text{'}=C_4{e}^x \\ {y}_p\text{'}\text{'}=C_4{e}^x \end{gathered}$$

Substitute the above derivatives,

$\begin{array}{rcl}y\text{'}\text{'}-2y\text{'}-3y& =& 4e^x-9\\ C_4{e}^x-2\left(C_4{e}^x\right)-3\left(C_3+C_4{e}^x\right)& =& 4e^x-9\\ -4C_4{e}^x-3C_3& =& 4e^x-9\end{array}$

Solve for general solution

Compare the coefficients of and constant terms in the above equation and solve for $C_3$and$C_4$ .

$C_3=3,C_4=-1$

Hence,$y_p=3-e^x$

$$\begin{gathered} y=y_c+y_p \\ y=C_1{e}^{-x}+C_2{e}^{3x}-e^x+3 \end{gathered}$$

Therefore, the general solution of the differential equation is $y=C_1{e}^{-x}+C_2{e}^{3x}-e^x+3$