In Problem 33 to 38, solve the given differential equations by using the principle of superposition [see the solution of equation (6.29)]. For example, in Problem 33, solve three differential equations with right-hand sides equal to the three different brackets. Note that terms with the same exponential factor are kept together; thus, a polynomial of any degree is kept together in one bracket.

${\mathbf{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{[}{\mathbf{x}}^3\mathbf{-}\mathbf{1}\mathbf{]}\mathbf{+}\mathbf{\left[}\mathbf{2}\mathbf{cosx}\mathbf{\right]}\mathbf{+}\mathbf{\left[}\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{)}{\mathbf{e}}^x\mathbf{\right]}$

The given equation is inhomogeneous equation. So, the solution of equation is as follows:

$y=y_e+y_p$

Where, $y_e$is the complementary function and $y_p$ is particular solution.

The auxiliary equation for is as follows:

$m^2+1=0$

Now we solve,

$\begin{array}{l}{m}^2+1=0\\ m^2=-1\\ m=\pm i\end{array}$

Since auxiliary equation$m^2+1=0$ has complex roots, the complementary function is as follows:

$\begin{array}{l}{y}_e=e^{0x}(C_1\cos x+C_2\sin x)\\ y_e=C_1\cos x+C_2\sin x\mathrm{...}\left(2\right)\end{array}$

In the given equation, right hand side is as follows:

$Q=[x^3-1]+\left[2\cos x\right]+\left[(2-4x)e^x\right]$

Let ,$Q=Q_1+Q_2+Q_3$were

The complete particular solution in this case is as follows:

$y_p=y_{p_1}+y_{p_2}+y_{p_3}$

For $Q_1=[x^3-1]$, calculate$y_{p_1}$for .$y^{\text{'}\text{'}}+y=[x^3-1]$

$\begin{array}{l}{y}_{p_1}=\frac{1}{P\left(D\right)}[x^3-1]\\ y_{p_1}=\frac{1}{(D^2+1)}[x^3-1]\\ y_{p_1}={(1+D^2)}^{-1}[x^3-1]\\ y_{p_1}=(1-D^2+D^4-\mathrm{...})[x^3-1]\end{array}$

Here, D represents first derivative of function of x, $D^2$represents second derivatives of function of x, and so on,

$\begin{array}{c}{y}_{p_1}=(1-D^2+D^4-\mathrm{...})[x^2-1]\\ =(x^3-1)-D^2(x^2-1)+D^4(x^3-1)-\mathrm{...}\\ =x^3-1-6x\end{array}$

For$Q_2=\left[2\cos x\right]$ , calculate $y_{p_2}$for $y^{\text{'}\text{'}}+y=\left[2\cos x\right]$

Since , $e^{ix}=\cos x+i\sin x$therefore $2\cos x=\mathrm{Re}\left[2e^{ix}\right]$

Calculate $y_{p_2}$for $y^{\text{'}\text{'}}+y=\mathrm{Re}\left[2e^{ix}\right]$

Consider $Y^{\text{'}\text{'}}+Y=2e^{ix}\mathrm{...}\left(4\right)$

Calculate$y=y_c+y_p$ for $Y^{\text{'}\text{'}}+Y=2e^{ix}$

For$Y^{\text{'}\text{'}}+Y=2e^{ix}$, $Y_p$is of the following form:

$Y_p=Cx{e}^{ix}$

Differentiate with respect to x.

$\begin{array}{c}{Y^{\text{'}}}_p=\frac{d}{dx}\left(Cx{e}^{ix}\right)\\ =Cxi{e}^{ix}+C{e}^{ix}\\ =C(xi{e}^{ix}+e^{ix})\end{array}$

Differentiate ${Y^{\text{'}}}_p$ with respect to x.

$\begin{array}{c}{Y^{\text{'}\text{'}}}_p=\frac{d}{dx}C(xi{e}^{ix}+e^{ix})\\ =C(x{i}^2{e}^{ix}+i{e}^{ix}+i{e}^{ix})\\ =C(-x{e}^{ix}+2i{e}^{ix})\end{array}$

Substitute ${Y^{\text{'}\text{'}}}_p=C(-x{e}^{ix}+2i{e}^{ix})$and$Y_p=Cx{e}^{ix}$in equation.

$\begin{array}{c}C(-x{e}^{ix}+2i{e}^{ix})+Cx{e}^{ix}=2e^{ix}\\ C{e}^{ix}(-x+2i+x)=2e^{ix}\\ C=\frac{2}{2i}\\ C=\frac{1}{i}\end{array}$

Simplifying further. We get,

$\begin{array}{c}C=\frac{1}{i}\times \frac{i}{i}\\ C=\frac{i}{i^2}\\ C=-i\end{array}$

Substitute$C=-i$ in equation,

$\begin{array}{c}{Y}_p=-ix{e}^{ix}\\ =-ix(\cos x+i\sin x)\\ =-ix\cos x+x\sin x\end{array}$

Now,

$\begin{array}{c}{y}_{p_2}=\mathrm{Re}\left[Y_p\right]\\ =\mathrm{Re}[-ix\cos x+x\sin x]\\ =x\sin x\mathrm{...}\left(6\right)\end{array}$

For , $Q_3=\left[(2-4x)e^x\right]$calculate$y_{p_3}$for $y^{\text{'}\text{'}}+y=\left[(2-4x)e^x\right]$

So, we get,

$\begin{array}{c}{y}_{p_3}=\frac{1}{D^2+1}\left((2-4x)e^x\right)\\ y_{p_3}=e^x\frac{1}{{(D+1)}^2+1}(2-4x)\\ y_{p_3}=e^x\frac{1}{D^2+2D+1+1}(2-4x)\\ y_{p_3}=e^x\frac{1}{D^2+2D+2}(2-4x)\end{array}$

Further simplify the above – mentioned equation,

$\begin{array}{c}{y}_{p_3}=e^x\frac{1}{2[\frac{D^2}{2}+D+1]}(2-4x)\\ y_{p_3}=e^x\frac{1}{[1+(D+\frac{D^2}{2})]}(2-4x)\\ y_{p_3}=\frac{e^x}{2}{[1+(D+\frac{D^2}{2})]}^{-1}(2-4x)\\ y_{p_3}=\frac{e^x}{2}[1-(D+\frac{D^2}{2})+{(D+\frac{D^2}{2})}^2-\mathrm{...}](2-4x)\end{array}$

Simplifying further,

$\begin{array}{c}{y}_{p_3}=\frac{e^x}{2}[(2-4x)-(D+\frac{D^2}{2})(2-4x)+{(D+\frac{D^2}{2})}^2(2-4x)-\mathrm{...}]\\ =\frac{e^x}{2}[(2-4x)-D(2-4x)-\frac{D^2}{2}(2-4x)+\mathrm{...}]\\ =\frac{e^x}{2}[(2-4x)+4+0]\\ =\frac{e^x}{2}[6-4x]\\ =e^x(3-2x)\mathrm{...}\left(7\right)\end{array}$

Substitute $y_{p_1}=x^3-1-6x,y_{p_2}=x\sin x,$and $y_p=y_{p_1}+y_{p_2}+y_{p_3}$

So,

$y_p=x^3-1-6x+x\sin x+e^x(3-2x)$

Substitute $y_e=C_1\cos x+C_2\sin x$and $y_p=x^3-1-6x+x\sin x+e^x(3-2x)$ in$y=y_e+y_p$ .

$\begin{array}{c}y=C_1\cos x+C_2\sin x+x^3-1-6x+x\sin x+e^x(3-2x)\\ =A\sin (x+\lambda )+x^3-1-6x+x\sin x+e^x(3-2x)\end{array}$

Where,

$\begin{array}{c}A\sin (x+\lambda )=A(\sin x\cos \lambda +\cos x\sin \lambda )\\ =A\sin x\cos \lambda +A\cos x\sin \lambda \\ =A\sin \lambda \cos x+A\cos \lambda \sin x\end{array}$

Therefore, the solution of$y^{\text{'}\text{'}}+y=[x^3-1]+\left[2\cos x\right]+\left[(2-4x)e^x\right]$ is

$\begin{array}{c}y=C_1\cos x+C_2\sin x+x^3-1-6x+x\sin x+e^x(3-2x)\\ y=A\sin (x+\lambda )+x^3-1-6x+x\sin x+e^x(3-2x)\end{array}$