Question

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

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

Short answer

$y=e^{-x}\left(C_1\cos x+C_2\sin x\right)+\frac{1}{10}{e}^{6x}$

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 $m^2+2m+2=0$we have $m=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\left(2\right)\left(1\right)}}{2\left(1\right)}=-1\pm i$

So, the complementary function is $y_c=e^{-x}\left(C_1\cos x+C_2\sin x\right)$.

Apply differential operator

From (5) of section we have $(D-6)e^{6x}=0$

So, $(D-6)\left(5e^{6x}\right)=0$

So, we apply $(D-6)$ to the both sides of the given differential equation gives

$\begin{array}{rcl}(D-6)\left(D^2+2D+2\right)y& =& (D-6)\left(5e^{6x}\right)\\ & =& 0\end{array}$

The auxiliary equation is

$$\begin{gathered} (m-6)\left(m^2+2m+2\right)=0 \\ \Rightarrow m_1=-1+i,m_2=-1-i\text{and}{m}_3=6 \end{gathered}$$

So, the general solution must be .$y=\underset{y_c}{\underbrace{e^{-x}\left(C_1\cos x+C_2\sin x\right)}}+C_3{e}^{6x}$

Solve for particular solution

Now particular solution,

$\begin{array}{rcl}{y}_p& =& A{e}^{6x}\\ & \Rightarrow & y_p\text{'}=6A{e}^{6x}\\ y_p\text{'}\text{'}& =& 36A{e}^{6x}\end{array}$

Substitution

Substituting these into the given differential equation gives

$\begin{array}{rcl}36A{e}^{6x}+12A{e}^{6x}+2A{e}^{6x}& =& 5e^{6x}\\ (50A-5)e^{6x}& =& 0\end{array}$

Since $e^{6x}\ne 0,50A-5=0$

$\therefore A=\frac{1}{10}$

So $y_p=\frac{1}{10}{e}^{6x}$

Thus, the general solution is $y=e^{-x}\left(C_1\cos x+C_2\sin x\right)+\frac{1}{10}{e}^{6x}$.