Question

In Problems 65–68 use a computer either as an aid in solving the auxiliary equation or as a means of directly obtaining the general solution of the given differential equation. If you use a CAS to obtain the general solution, simplify the output and, if necessary, write the solution in terms of real functions.

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

Short answer

$y=k_{1 }{e}^{\frac{1}{2}x} \cos \frac{\sqrt{3}}{2}x+k_{2 }{e}^{\frac{1}{2}x} \sin \frac{\sqrt{3}}{2}x+k_{3 }{e}^{-\frac{1}{2}x}\cos \frac{\sqrt{7}}{2}x+k_{4 }{e}^{-\frac{1}{2}x}\sin \frac{\sqrt{7}}{2}x$

Step-by-step solution

Definition of general equation

We previously concluded that the given system's is general solution then its used the type of matrix form.

The third order differential equation:

$y^{\left(4\right)}+2y\text{'}\text{'}-y\text{'}+2y=0$

substitute$y=e^{mx},y\text{'}=m{e}^{mx}$ and$y\text{'}\text{'}=m^2{e}^{mx},y\text{'}\text{'}\text{'}=m^3{e}^{mx}\mathrm{and}{y}^{\left(4\right)}=m^4{e}^{mx}$ into$y^{\left(4\right)}+2y\text{'}\text{'}-y\text{'}+2y=0$ to obtain the auxiliary equation.

$\begin{array}{c}{m}^4 e^{mx}+2m^2 e^{mx}-m{e}^{mx}+2 e^{mx}=0\\ e^{mx}\left(m^4+2m^2-m+2\right)=0\end{array}$

For any real,$x,e^{mx}\ne 0$ we have

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

Auxiliary Equation of the given differential equation.

This auxiliary equation cannot be solved easily, then using Computer Algebra System we can solve it and obtain the roots as

$m_1=\frac{1}{2}+\frac{\sqrt{3}}{2}i, m_2=\frac{1}{2}-\frac{\sqrt{3}}{2}i, m_3=-\frac{1}{2}+\frac{\sqrt{7}}{2}i, m_4-=-\frac{1}{2}-\frac{\sqrt{7}}{2}i$

Find the general solution of the homogeneous equation

Homogeneous Linear Second Order Equation is a linear combination ofare real and distinct.

which are mix of real and conjugate complex Then the general solution of this homogeneous equation can be given.

$\begin{array}{l}{y}^{\left(4\right)}+2y\text{'}\text{'}-y\text{'}+2y=0\\ y=c_{1 } e^{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)x}+c_2 e^{\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)x}+c_3 e^{\left(-\frac{1}{2}+\frac{\sqrt{7}}{2}i\right)x}+c_4 e^{\left(-\frac{1}{2}-\frac{\sqrt{7}}{2}i\right)x}\\ y=c_{2 }{e}^{\frac{1}{2}x}\left(\cos \frac{\sqrt{3}}{2}x+\sin \frac{\sqrt{3}}{2}x\right)+c_{2 }{e}^{\frac{1}{2}x}\left(\cos \frac{\sqrt{3}}{2}x-\sin \frac{\sqrt{3}}{2}x\right)+c_{3 }{e}^{-\frac{1}{2}x}\left(\cos \frac{\sqrt{7}}{2}x+\sin \frac{\sqrt{7}}{2}x\right)+c_{4 }{e}^{-\frac{1}{2}x}\left(\cos \frac{\sqrt{7}}{2}x-\sin \frac{\sqrt{7}}{2}x\right)\\ y=\left(c_{1 }+c_{2 }\right)e^{\frac{1}{2}x}\cos \frac{\sqrt{3}}{2}x+\left(c_{1 }-c_{2 }\right)e^{\frac{1}{2}x}\sin \frac{\sqrt{3}}{2}x+\left(c_{3 }+c_{4 }\right)e^{-\frac{1}{2}x}\cos \frac{\sqrt{7}}{2}x+\left(c_{3 }-c_{4 }\right)e^{-\frac{1}{2}x}\sin \frac{\sqrt{7}}{2}x\\ y=k_{1 }{e}^{\frac{1}{2}x} \cos \frac{\sqrt{3}}{2}x+k_{2 }{e}^{\frac{1}{2}x} \sin \frac{\sqrt{3}}{2}x+k_{3 }{e}^{-\frac{1}{2}x}\cos \frac{\sqrt{7}}{2}x+k_{4 }{e}^{-\frac{1}{2}x}\sin \frac{\sqrt{7}}{2}x\end{array}$

Arbitrary constants:

$k_{1 }=c_{1 ,}{k}_{2 }=c_{2 },k_{3 }=\left(c_{3 }+c_{4 }\right)\mathrm{and}\left(c_{3 }-c_{4 }\right)$

Hence the solution is

$y=k_{1 }{e}^{\frac{1}{2}x} \cos \frac{\sqrt{3}}{2}x+k_{2 }{e}^{\frac{1}{2}x} \sin \frac{\sqrt{3}}{2}x+k_{3 }{e}^{-\frac{1}{2}x}\cos \frac{\sqrt{7}}{2}x+k_{4 }{e}^{-\frac{1}{2}x}\sin \frac{\sqrt{7}}{2}x$