Question

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

$\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}{\left(e^x+e^{-x}\right)}^{\mathbf{2}}$

Short answer

$y=C_1{e}^{\frac{x}{2}}+C_2{e}^{2x}+C_3{e}^{-x}+1+\frac{1}{9}x{e}^{2x}-\frac{1}{20}{e}^{-2x}$

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 the differential equation:$2y\text{'}\text{'}\text{'}-3y\text{'}\text{'}-3y\text{'}+2y={\left(e^x+e^{-x}\right)}^2$

Put$y\text{'}=m,y\text{'}\text{'}=m^2 \& y\text{'}\text{'}\text{'}=m^3$ we have:

$2m^3-3m^2-3m+2=0$.

By solving the equation we have:

$m_1=-1,m_2=\frac{1}{2}$, and$m_3=2$

So, the complementary function is $y_c=C_1{e}^{\frac{x}{2}}+C_2{e}^{2x}+C_3{e}^{-x}$[ As$m_1,m_2 \& m_3$ are all different]

Form differential equation

Also${\left(e^x+e^{-x}\right)}^2=e^{2x}+e^{-2x}+2$

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

So,$D(D-2)(D+2)\left[{\left(e^x+e^{-x}\right)}^2\right]=0$

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

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

i.e,$D(D-2{)}^2(D+2)(D+1)\left(D-\frac{1}{2}\right)y=0\cdots \left(1\right)$

General solution

The auxiliary equation for equationis:

$$\begin{gathered} m(m-2)(m+2)(m+1)\left(m-\frac{1}{2}\right)=0 \\ \Rightarrow m_1=0,m_2=-2,m_3=2=m_4,m_5=-1\text{and}{m}_6=\frac{1}{2} \end{gathered}$$

So, the general solution is$y=\underset{y_c}{\underbrace{C_1{e}^{\frac{x}{2}}+C_2{e}^{2x}+C_3{e}^{-x}}}+C_4+C_{5}x{e}^{2x}+C_6{e}^{-2x}$

Now, particular solution of the given differential equation should also satisfy and the particular solution is given by:$y_p=A+Bx{e}^{2x}+C{e}^{-2x}$ corresponding to last three terms in general solution.

Find particular solution

We have $y_p=A+Bx{e}^{2x}+C{e}^{-2x}$differentiating we get:

$$\begin{gathered} y_p\text{'}=2Bx{e}^{2x}+B{e}^{2x}-2C{e}^{-2x} \\ {\dot{y}}_p\text{'}\text{'}=4Bx{e}^{2x}+4C{e}^{-2x}+4B{e}^{2x} \\ {y}_p\text{'}\text{'}\text{'}=8Bx{e}^{2x}-8C{e}^{-2x}+12B{e}^{2x} \end{gathered}$$

Substituting these values in the given differential equation gives

$\begin{array}{rcl}-20C{e}^{-2x}+9B{e}^{2x}+2A& =& {\left(e^x+e^{-x}\right)}^2\\ & =& e^{-2x}+e^{2x}+2\end{array}$

Equating the coefficients of like terms, we get

$$\begin{gathered} -20C=1 \\ 9B=1 \\ 2A=2 \end{gathered}$$

$A=1,B=\frac{1}{9} and C=-\frac{1}{20}$

So, the particular solution is.$y_p=1+\frac{1}{9}x{e}^{2x}-\frac{1}{20}{e}^{-2x}$

Thus, the general solution of the given differential equation is $y=C_1{e}^{\frac{x}{2}}+C_2{e}^{2x}+C_3{e}^{-x}+1+\frac{1}{9}x{e}^{2x}-\frac{1}{20}{e}^{-2x}$