Question

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

$\mathbf{y}\text{'}\text{'}\mathbf{-}\mathbf{y}\text{'}\mathbf{-}\mathbf{12}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{x}}$

Short answer

$y=C_1{e}^{-3x}+C_2{e}^{4x}+\frac{1}{7}x{e}^{4x}$

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

From$m^2-m-12=0$we find$m_1=-3,m_2=4$

So, the complementary function is $y_c=C_1{e}^{-3x}+C_2{e}^{4x}$.

 Apply differential operator

Since$(D-4)e^{4x}=0$

Now we apply$(D-4)$to the both sides of the given differential equation gives

$\begin{array}{rcl}(D-4)\left(D^2-D-12\right)y& =& (D-4)\left(e^{4x}\right)\\ & =& 0\end{array}$

The auxiliary equation is

$\begin{array}{rcl}(m-4)\left(m^2-m-12\right)& =& 0\\ & \Rightarrow & m_1=-3,m_2=4=m_3\end{array}$

So, the general solution is $y=\underset{y_c}{\underbrace{C_1{e}^{-3x}+C_2{e}^{4x}+C_{3}x{e}^{4x}}}$.

Solve for particular Solution

Now particular solution,$y_p$

$y_p=Ax{e}^{4x}$

Then$y_p\text{'}=A{e}^{4x}+4Ax{e}^{4x}$

$y_p\text{'}\text{'}=8A{e}^{4x}+16Ax{e}^{4x}$

Substitution

Now, substituting these into the given differential equation gives

$\begin{array}{rcl}8A{e}^{4x}+16Ax{e}^{4x}-A{e}^{4x}-4Ax{e}^{4x}-12Ax{e}^{4x}& =& e^{4x}\\ 7A{e}^{4x}& =& e^{4x}\\ (7A-1)e^{4x}& =& 0\end{array}$

Since$e^{4x}\ne 0$ , we have$7A-1=0$

$A=\frac{1}{7}$

So$y_p=\frac{1}{7}x{e}^{4x}$

Thus, the general solution is .role="math" localid="1663914337573" $y=C_1{e}^{-3x}+C_2{e}^{4x}+\frac{1}{7}x{e}^{4x}$