Question

Solve the given initial-value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{5}\mathbf{+}\mathbf{8}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{4}$

Short answer

$y=\left(-\frac{177}{72}\cos \sqrt{3x}+\frac{17\sqrt{3}}{72}\sin \sqrt{3x}\right)e^x-\frac{138}{72}{e}^{-2x}+\frac{1}{4}x-\frac{5}{8}+\frac{2}{3}x{e}^{-2x}$

Step-by-step solution

Obtain the general solution of the homogeneous differential equation   

First, we need to solve the associated homogeneous equation $y\text{'}\text{'}\text{'}+8y=0$

Corresponding auxiliary equation is$m^3+8=0$

$(m^2-2m+4)+(m+2)=0$

Solving,

$m_1=1+\sqrt{3}i$, $m_2=1-\sqrt{3}i$ and $m_3=-2$ , which are combination of real and complex

So the solution of corresponding homogeneous equation is

$m_3=-2m_3=-2$

Simplifying

$y_h=e^x(k_1\cos \sqrt{3x}+k_2\sin \sqrt{3x})+k_3{e}^{-2x}$

Finding the particular solution 

Given non homogeneous differential equation is$y\text{'}\text{'}\text{'}+8y=2x-5+8e^{-2x},$

Assume that $y_p=Ax+B+Cx{e}^{-2x}······\left(1\right)$is a solution of the non-homogeneous differential equation

Differentiate with respect to x

$y_p\text{'}=A+C{e}^{-2x}-2Cx{e}^{-2x}······\left(2\right)$

role="math" localid="1664202795014" $$\begin{gathered} y_p\text{'}\text{'}=-2C{e}^{-2x}-2C{e}^{-2x}+4Cx{e}^{-2x} \\ =-4C{e}^{-2x}+4Cx{e}^{-2x}······\left(3\right) \end{gathered}$$

$$\begin{gathered} y_p\text{'}\text{'}\text{'}=8C{e}^{-2x}+4C{e}^{-2x}-8Cx{e}^{-2x} \\ =12C{e}^{-2x}-8Cx{e}^{-2x}······\left(4\right) \end{gathered}$$

Substituting (1) and (4) in given differential equation

$(12C{e}^{-2x}-8Cx{e}^{-2x})+8[Ax+B+Cx{e}^{-2x}]=2x-5+8e^{-2x}$

Or, $8Ax+8B+12C{e}^{-2x}=2X-5+8e^{-2x}$

comparing the coefficients,

$8A=2\Rightarrow A=\frac{1}{4}$

$8B=-5\Rightarrow B=\frac{-5}{8}$

$12C=8\Rightarrow C=\frac{2}{3}$

Hence the particular solution is

$y_p=\frac{1}{4}x-\frac{5}{8}+\frac{2}{3}x{e}^{-2x}$

Finding the solution

Hence the solution of given differential equation is

y= y_h + y_p

$y=e^x(k_1\cos \sqrt{3x}+k_2\sin \sqrt{3x})+k_3{e}^{-2x}+\frac{1}{4}x-\frac{5}{8}+\frac{2}{3}x{e}^{-2x}······\left(5\right)$

Applying the given conditions  

At the point (0,-5)

(5) becomes

$-5=e^0(k_1\cos 0+k_2\sin 0)+k_3{e}^0+\frac{1}{4}\left(0\right)-\frac{5}{8}+\frac{2}{3}\left(0\right)$

Or $k_1+k_3=-\frac{35}{8}$

Finding derivative of (5)

$$\begin{gathered} y\text{'}=e^x(k_1\cos \sqrt{3x}+k_2\sin \sqrt{3x})+e^x(-\sqrt{3}{k}_1\sin \sqrt{3x}+\sqrt{3}{k}_2\cos \sqrt{3x})-2k_3{e}^{-2x}+\frac{1}{4}+ \\ \frac{2}{3}{e}^{-2x}-\frac{4}{3}x{e}^{-2x} \end{gathered}$$

Applying the given condition, at (0,3)

$3=e^0(k_1\cos 0+k_2\sin 0)+e^0(-\sqrt{3}{k}_1\sin 0+\sqrt{3}{k}_2\cos 0)-2k_3{e}^0+\frac{1}{4}+\frac{2}{3}{e}^0$

Finding second derivative of (5)

$y\text{'}\text{'}=e^x\left[\right(-2k_1+2\sqrt{3}{k}_2\left)\cos \sqrt{3}x\right)+(-2\sqrt{3}{k}_1-2k_2)\sin \sqrt{3}x\left)\right]+4k_3{e}^{-2x}-\frac{8}{3}{e}^{-2x}+\frac{8}{3}x{e}^{-2x}$

At (0,-4)

$-4=e^0\left[\right(-2k_1+2\sqrt{3}{k}_2\left)\cos 0\right)+(-2\sqrt{3}{k}_1-2k_2)\sin 0\left)\right]+4k_3{e}^0-\frac{8}{3}{e}^0+\frac{8}{3}\left(0\right)e^0$

$2k_1+2\sqrt{3}{k}_2+4k_3=-\frac{4}{3}$

Solving,$k_1=-\frac{177}{72}$ ,$k_2=\frac{17\sqrt{3}}{72}$ ,$k_3=-\frac{138}{72}$

Hence the solution is $\mathit{y}\mathbf{=}\left(-\frac{177}{72}cos\sqrt{3x}+\frac{17\sqrt{3}}{72}sin\sqrt{3x}\right){\mathit{e}}^{\mathbf{x}}\mathbf{-}\frac{\mathbf{138}}{\mathbf{72}}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathit{x}\mathbf{-}\frac{\mathbf{5}}{\mathbf{8}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}$