Question

Given that $\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{x}$ is a solution of ${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{11}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{10}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{\text{,}}$

find the general solution of the DE without the aid of a calculator or a computer.

Short answer

The general solution of the DE is$y=c_1\cos x+c_2\sin x+e^{-x}\left(c_3\cos 3x+c_4\sin 3x\right)$

Step-by-step solution

Define the general solution of the DE.

A general solution to an nth order differential equation involves n arbitrary constants. If we use the variables separable technique to solve a first order differential equation, we must include an arbitrary constant as soon as the integration is completed.

Now calculate the general solution.

we have to assume that $y=e^{mx}$is a solution of the given D.E,

then differentiate with respect to then we have

$$\begin{gathered} y^{\text{'}}=m{e}^{mx} \\ {y}^{\text{'}\text{'}}=m^2{e}^{mx} \\ {y}^{\text{'}\text{'}\text{'}}=m^3{e}^{mx} \\ {y}^{\left(4\right)}=m^4{e}^{mx} \end{gathered}$$

Then substitute with our assumption $y=e^{mx}$and with these derivatives into the given differential equation, then we have

$$\begin{gathered} m^4{e}^{mx}+2m^3{e}^{mx}+11m^2{e}^{mx}+2m{e}^{mx}+10e^{mx}=0 \\ \left(m^4+2m^3+11m^2+2m+10\right)e^{mx}=0 \end{gathered}$$

Since$e^{mx}$ can not be 0 ,

$m^4+2m^3+11m^2+2m+10=0$

make a long division for the auxiliary equation on this factor to find the remain solutions as $m^4+2m^3+11m^2+2m+10=\left(m^2+1\right)\left(m^2+2m+10\right)=0$

Then the roots are

$$\begin{gathered} m_{1,2}=\pm i \\ {m}_{3,4}=\frac{-2\pm \sqrt{4-4\times 1\times 10}}{2} \\ =\frac{-2\pm \sqrt{-36}}{2}=-1\pm 3i \end{gathered}$$

$$\begin{gathered} y=c_1\cos x+c_2\sin x+c_3{e}^{(-1+3i)x}+c_4{e}^{(-1-3i)x} \\ =c_1\cos x+c_2\sin x+e^{-x}\left(c_3\cos 3x+c_4\sin 3x\right) \end{gathered}$$

The general solution is$y=c_1\cos x+c_2\sin x+e^{-x}\left(c_3\cos 3x+c_4\sin 3x\right)$.