Question

In Problemssolve the given initial-value problem.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{8}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{3}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\left(0\right)\mathbf{=}\mathbf{2}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{4}$

Short answer

$y=2e^{2x}\cos 2x-\frac{3}{64}{e}^{2x}\sin 2x+\frac{1}{8}{x}^3+\frac{3}{16}{x}^2+\frac{3}{32}x$

Step-by-step solution

Definition of initial value problem

An ordinary differential equation in which an initial condition is provided that identifies the unknown function within the domain is known as the initial value problem.

Complementary function

Consider the differential$y^{\text{'}\text{'}}-4y^{\text{'}}+8y=x^3,y\left(0\right)=2,y^{\text{'}}\left(0\right)=4$ .

The auxiliary equation corresponding to homogenous equation gives:

$m^2-4m+8=0$

we find:

$\begin{array}{rcl}m& =& \frac{4\pm \sqrt{{\left(-4\right)}^2-4\left(1\right)\left(8\right)}}{2\left(1\right)}\\ & =& 2\pm 2i\end{array}$

So, the complementary function is $y_c=e^{2x}\left(C_1\cos 2x+C_2\sin 2x\right)$.

General solution

As , $D^4\left(x^3\right)=0D^4\left(x^3\right)=0$we apply the differential operator $D^4$to the both sides of the given differential equation. So, we get

$\begin{array}{rcl}{D}^4\left(D^2-4D+8\right)y& =& D^4\left(x^3\right)\\ D^4\left(D^2-4D+8\right)y& =& 0\end{array}$

The auxiliary equation is$m^4\left(m^2-4m+8\right)=0$

$\begin{array}{rcl}& \Rightarrow & m_1=2+2i,m_2=2-2i\\ m_3& =& 0=m_4=m_5=m_6\end{array}$

So, the general solution is: .$y=\underset{y_c}{\underbrace{e^{2x}\left(C_1\cos 2x+C_2\sin 2x\right)}}+C_3+C_{4}x+C_5{x}^2+C_6{x}^3$

Particular solution

The 4 terms denotes particular solution so equation is written as:

$y_p=A+Bx+C{x}^2+D{x}^3$

Then

$\begin{array}{rcl}{y}_p^{\text{'}}& =& B+2Cx+3D{x}^2\\ y_p^{\text{'}\text{'}}& =& 2C+6Dx\end{array}$

So, the given differential equation gives

$\begin{array}{rcl}\left(2C+6Dx\right)-4\left(B+2Cx+3D{x}^2\right)+8\left(A+Bx+C{x}^2+D{x}^3\right)& =& x^3\\ 8D{x}^3+\left(8C-12D\right)x^2+\left(6D+8B-8C\right)x+\left(8A-4B+2C\right)& =& x^3\end{array}$

On equating the coefficients of like terms& solving for coefficients we have:

$D=\frac{1}{8},C=\frac{3}{16},B=\frac{3}{32},A=0$

So,$y_p=\frac{3}{32}x+\frac{3}{16}{x}^2+\frac{1}{8}{x}^3$

Thus, the general solution is$y=e^{2x}\left(C_1\cos 2x+C_2\sin 2x\right)+\frac{3}{32}x+\frac{3}{16}{x}^2+\frac{1}{8}{x}^3$ .

Apply initial condition

$y^{\text{'}}\left(0\right)=4$The value of$y^{\text{'}}\left(x\right)$ is

$y^{\text{'}}\left(x\right)=2e^{2x}\left(C_1\cos 2x+C_2\sin 2x\right)+e^{2x}\left(-2C_1\sin 2x+2C_2\cos 2x\right)+\frac{3}{32}+\frac{3}{8}x+\frac{3}{8}{x}^2$

Put ,$y\left(0\right)=2$we get$2=C_1$

And setting$y^{\text{'}}\left(0\right)=4$ , we get$4=2\left(2\right)+2C_2+\frac{3}{32}$

$C_2=-\frac{3}{64}$

Thus, the solution of the given initial-value problems is$y=2e^{2x}\cos 2x-\frac{3}{64}{e}^{2x}\sin 2x+\frac{1}{8}{x}^3+\frac{3}{16}{x}^2+\frac{3}{32}x$