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.

$3.15y^{\left(4\right)}-5.34y\text{'}\text{'}+6.33y\text{'}-2.03y=0$

Short answer

$y=k_1{e}^{0.5012x}+k_2{e}^{-1.748x}+k_3{e}^{-0.623x}cos0.589x+k_4{e}^{-0.623x}sin0.589x$

Step-by-step solution

Find solution for homogeneous differential equation

We have the third order differential equation

$3.15y^{\left(4\right)}-5.34y\text{'}\text{'}+6.33y\text{'}-2.03y=0$

Consider$\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{m}\mathbf{x}}$as the solution of the differential equation,

Substitute$y=e^{mx},y\text{'}=m{e}^{mx}$ and$y\text{'}\text{'}=m^2{e}^{mx},y\text{'}\text{'}\text{'}=m^3{e}^{mx}\mathrm{and}{y}^{\left(4\right)}=m^4{e}^{mx}$ intorole="math" localid="1663915471361" $3.15y^{\left(4\right)}-5.34y\text{'}\text{'}+6.33y\text{'}-2.03y=0$ to obtain the auxiliary equation.

role="math" localid="1663915617663" $\begin{array}{c}3.15m^4{e}^{mx}-5.34m^2{e}^{mx}+6.33m{e}^{mx}-2.03e^{mx}=0\\ e^{mx}\left(3.15m^4-5.34m^2+6.33m-2.03\right)=0\end{array}$

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

role="math" localid="1663915632372" $3.15m^4-5.34m^2+6.33m-2.03=0$

Is the auxiliary equation of the given differential equation.

Use computer algebra system

The auxiliary equation cannot be solved easily, then using computer algebra system we can solve it and obtain the root as

$m_1=0.5012,m_2=-1.748,m_3=-0623+0.589iand{m}_4=-0.6230.589i$

Which are mix of real and conjugate complex

Then the general solution of this homogeneous equation

$3.15y^{\left(4\right)}-5.34y\text{'}\text{'}+6.33y\text{'}-2.03y=0$

can be given as

$\mathrm{y}=c_1{e}^{0.5012x}+c_2{e}^{-1.748x}+c_3{e}^{\left(-0623+0.589i\right)x}+c_4{e}^{(-0.0623-0.589i)x}$

Then we have

role="math" localid="1663915721832" $\mathrm{y}=c_1{e}^{0.5012x}+c_2{e}^{-1.748x}+c_3{e}^{-0.623x}\left(cos0.589x+sin0.589x\right)+c_4{e}^{-0.623x}\left(cos0.589x-sin0.589x\right)$

Then we have

$\mathrm{y}=c_1{e}^{0.5012x}+c_2{e}^{-1.748x}+\left(c_3+c_4\right)e^{-0.623x}cos0.589x+\left(c_3-c_4\right)e^{-0.623x}sin0.589x$

Then we obtain

$y=k_1{e}^{0.5012x}+k_2{e}^{-1.748x}+k_3{e}^{-0.623x}cos0.589x+k_4{e}^{-0.623x}sin0.589x$

Where $k_1=c_1,k_2=c_2,k_3=\left(c_3+c_4\right)\mathrm{and}{k}_4=\left(c_3-c_4\right)$ are arbitrary constants.

Hence the final solution is

$y=k_1{e}^{0.5012x}+k_2{e}^{-1.748x}+k_3{e}^{-0.623x}cos0.589x+k_4{e}^{-0.623x}sin0.589x$