Question

Solve the given differential equation by undetermined coefficients.

${\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{6}{\mathbf{y}}^{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{-}{\mathbf{xe}}^{\mathbf{4}\mathbf{x}}$

Short answer

$y=C_1{e}^{-3x}+C_{2}x{e}^{-3x}+\frac{2}{343}{e}^{4x}-\frac{1}{49}x{e}^{4x}$

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

We are given a function $y^{\text{'}\text{'}}+6y^{\text{'}}+9y=-x{e}^{4x}$.

The objective is to solve the above equation using the method of undetermined coefficients.

Evaluation

The auxiliary equation of$y^{\text{'}\text{'}}+6y^{\text{'}}+9y=0$ corresponding to (1) is,

$$\begin{gathered} m^2+6m+9=0 \\ (m+3{)}^2=0 \\ m=-3,m=-3 \end{gathered}$$

Then, the complementary function is,

$$\begin{gathered} y_c=C_1{e}^{mx}+C_{2}x{e}^{mx} \\ {y}_c=C_1{e}^{-3x}+C_{2}x{e}^{-3x} \end{gathered}$$

Apply differential factor

Since the coefficient of in the power of$e^{4x}$ is 4 ,

$(D-4{)}^2\left(-x{e}^{4x}\right)=0$

Now, apply $(D-4{)}^2$to the both sides of the differential equation (1).

$$\begin{gathered} (D-4{)}^2\left(D^2+6D+9\right)y=(D-4{)}^2\left(-x{e}^{4x}\right) \\ (D-4{)}^2\left(D^2+6D+9\right)y=0\dots \dots \left(2\right) \end{gathered}$$

Find particular solution

Now, the auxiliary equation corresponding to (2) is,

$$\begin{gathered} (m-4{)}^2\left(m^2+6m+9\right)=0 \\ {m}_1=-3=m_2,m_3=4=m_4 \end{gathered}$$

So, the general solution must be of the form,

$y=\underset{y_c}{\underbrace{C_1{e}^{-3x}+C_{2}x{e}^{-3x}}}+C_3{e}^{4x}+C_{4}x{e}^{4x}$

The particular solution $y_p$should also satisfy (2) and note that the first two terms in y are attributed by $y_c$.

Basic form

So, the remaining terms in y must be the basic form of$y_p$ :

$y_p=A{e}^{4x}+Bx{e}^{4x}$

Then,

$$\begin{gathered} y_p^{\text{'}}=4A{e}^{4x}+B{e}^{4x}+4Bx{e}^{4x} \\ {y}_p^{\text{'}\text{'}}=16A{e}^{4x}+4B{e}^{4x}+4B{e}^{4x}+16Bx{e}^{4x} \\ {y}_p^{\text{'}\text{'}}=16A{e}^{4x}+8B{e}^{4x}+16Bx{e}^{4x} \end{gathered}$$

Substitution

Substitute these into the differential equation (1).

$\begin{array}{rcl}{y}_p^{\text{'}\text{'}}+6y_p^{\text{'}}+9y_p& =& -x{e}^{4x}\\ \left(16A{e}^{4x}+8B{e}^{4x}+16Bx{e}^{4x}\right)+6\left(4A{e}^{4x}+B{e}^{4x}+4Bx{e}^{4x}\right)+9\left(A{e}^{4x}+Bx{e}^{4x}\right)& =& -x{e}^{4x}\\ 16A{e}^{4x}+8B{e}^{4x}+16Bx{e}^{4x}+24A{e}^{4x}+6B{e}^{4x}+24Bx{e}^{4x}+9A{e}^{4x}+9Bx{e}^{4x}& =& -x{e}^{4x}\\ 49Bx{e}^{4x}+(49A+14B)e^{4x}& =& -x{e}^{4x}\end{array}$

Equate the coefficient

Equate the coefficients of$x{e}^{4x},e^{4x}$ on both sides of equation.

$$\begin{gathered} 49B=-1 \\ 49A+14B=0 \end{gathered}$$

Then,

$\begin{array}{rcl}B& =& -\frac{1}{49}\\ 49A+14\left(-\frac{1}{49}\right)& =& 0\\ 49A+2\left(-\frac{1}{7}\right)& =& 0\\ 49A& =& \frac{2}{7}\\ A& =& \frac{2}{343}\end{array}$

Put ,$A=\frac{2}{343}$ and$B=-\frac{1}{49}$ in .$y_p=A{e}^{4x}+Bx{e}^{4x}$

$y_p=\frac{2}{343}{e}^{4x}-\frac{1}{49}x{e}^{4x}$

Thus, the general solution of the given differential equation is,$y=C_1{e}^{-3x}+C_{2}x{e}^{-3x}+\frac{2}{343}{e}^{4x}-\frac{1}{49}x{e}^{4x}$

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