Question

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

${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{5}{\mathit{x}}^{\mathbf{2}}\mathbf{-}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$

Short answer

$y=C_1{e}^{-2x}+C_2{e}^{2x}+C_3+C_{4}x-\frac{5}{16}{x}^2-\frac{5}{48}{x}^4-\frac{1}{16}x{e}^{2x}$

Step-by-step solution

Definition

The method of undetermined coefficients is an approach to finding a particular solution to certain non homogeneous ordinary differential equations and recurrence relations

Form complementary solution

Consider differential Equation is$y^{\left(4\right)}-4y\text{'}\text{'}=5x^2-e^{2x}$

The homogeneous Equation is$y^{\left(4\right)}-4y\text{'}\text{'}=0$

Auxiliary equation is$m^4-4m^2=0$

$\Rightarrow m_1=2,m_2=-2$and$m_3=0=m_4$

So, the complementary function is$y_c=C_1{e}^{-2x}+C_2{e}^{2x}+C_3+C_{4}x$

General solution

Notice that$D^3\left(x^2\right)=0$

So $(D-\alpha {)}^n$annihilates each of the functions$e^{ax},x{e}^{ax},x^2{e}^{ax},\dots ,x^{n-1}{e}^{ax}$

We have$(D-2)e^{2x}=0$

$D^3(D-2)\left[5x^2-e^{2x}\right]=0$

Applying$D^3(D-2)$to the both sides of the given differential equation gives

$\begin{array}{rcl}{D}^3(D-2)\left(D^4-4D^2\right)y& =& D^3(D-2)\left[5x^2-e^{2x}\right]\\ D^5(D-2{)}^2(D+2)y& =& 0 \end{array}$

$\begin{array}{rcl}{m}^5(m-2{)}^2(m+2)& =& 0\\ & \Rightarrow & m_1=2=m_2,m_3=-2,m_4=0=m_5=m_6=m_7=m_8\end{array}$

Thus, the general solution must be$y=\underset{y_c}{\underbrace{C_1{e}^{-2x}+C_2{e}^{2x}+C_3+C_{4}x}}+C_{5}x{e}^{2x}+C_6{x}^2+C_7{x}^3+C_8{x}^4$

Particular solution

The last 4 terms in general solution gives particular solution & its is in the form:

$y_p=A{x}^2+B{x}^3+C{x}^4+Dx{e}^{2x}$

Differentiating we have:

$\begin{array}{rcl}{y}_p\text{'}& =& 2Ax+3B{x}^2+4C{x}^3+D{e}^{2x}+2Dx{e}^{2x}\\ y_p\text{'}\text{'}& =& 2A+6Bx+12C{x}^2+4D{e}^{2x}+4Dx{e}^{2x}\\ y_p\text{'}\text{'}\text{'}& =& 6B+24Cx+8D{e}^{2x}x+12D{e}^{2x}\\ y_p^{\left(4\right)}& =& 24C+32D{e}^{2x}+16Dx{e}^{2x}\end{array}$

Substituting$y_p$into the given differential equation gives

$\begin{array}{rcl}24C+32D{e}^{2x}+16Dx{e}^{2x}-8A-24Bx-48C{x}^2-16D{e}^{2x}-16Dx{e}^{2x}& =& 5x^2-e^{2x}\\ 16D{e}^{2x}-24Bx-48C{x}^2+(24C-8A)& =& 5x^2-e^{2x}\end{array}$

On equating the coefficients of like terms, we get

$\begin{array}{rcl}-8A+24C& =& 0\\ -24B& =& 0\\ -48C& =& 5\\ 16D& =& -1\end{array}$

Then$A=-\frac{5}{16},B=0,C=-\frac{5}{48} and D=-\frac{1}{16}$

So,$y_p=-\frac{5}{16}{x}^2-\frac{5}{48}{x}^4-\frac{1}{16}x{e}^{2x}$

Thus, the general solution of the given differential equation is $y=C_1{e}^{-2x}+C_2{e}^{2x}+C_3+C_{4}x-\frac{5}{16}{x}^2-\frac{5}{48}{x}^4-\frac{1}{16}x{e}^{2x}$