Question

Verify that$y=\sinh x-2 \cos \left(x+\frac{\pi }{6}\right)$ is a particular solution of$y^{\left(4\right)}-y=0$ . Reconcile this particular solution with the general solution of the DE.

Short answer

$\begin{array}{l}{k}_1=\frac{1}{2}, k_2=-\frac{1}{2}, k_3=-\sqrt{3} k_4=1\\ y =\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x}-\sqrt{3}\cos x+\sin x x\end{array}$

Step-by-step solution

General solution

The general solution of the DE is

$y=c_1{e}^x+c_2{e}^{-x}+c_3{e}^{i x}+c_4{e}^{-i x}$

Solving the auxiliary equation

To verify that$y=\sinh x-2 \cos \left(x+\frac{\pi }{6}\right)$is a particular solution for the differential equation$y^{\left(4\right)}-y=0$,

Consider $y=e^{mx}$as the solution of the differential equation,

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

$\begin{array}{c}{m}^4 e{}^{mx}- e{}^{mx}=0\\ e{}^{mx}\left(m^4-1\right)=0\end{array}$

Since$e^{mx}$cannot equal zero any more, then we can have

$\begin{array}{c} m^4-1=0\\ \left(m^2-1\right) \left(m^2+1\right)=0\\ \left(m-1\right) \left(m+1\right)\left(m^2+1\right)=0\end{array}$

Then the roots are

$m_1=1 m_2=-1\mathrm{and}{m}_{3,4}=\pm i$

which are mix of real and complex.

Then we obtain the general solution of our differential equationas

role="math" localid="1663911157106" $\begin{array}{l}y=c_1{e}^x+c_2{e}^{-x}+c_3{e}^{i x}+c_4{e}^{-i x}\\ =c_1{e}^x+c_2{e}^{-x}+c_3\left(\cos x+\sin x\right)+c_4\left(\cos x-\sin x\right)\\ =k_1{e}^x+k_2{e}^{-x}+k_3 \cos x+k_4\sin x ······\left(\mathrm{a}\right)\end{array}$

where we have$k_1=c_1, k_2=c_2, k_3=c_3+c_4\mathrm{and}{k}_4=c_3-c_4$ are arbitrary constants.

Simplify the equation

Second, we have to simplify the alleged solution as

$\begin{array}{l}y=\sinh x-2 \cos \left(x+\frac{\pi }{6}\right) \\ =\frac{e^x-e^{-x}}{2}-2\left(\cos x \sin \left(\frac{\pi }{6}\right) - \sin x \cos \left(\frac{\pi }{6}\right)\right)\\ =\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x}-2\left(\cos x \times \left(\frac{\sqrt{3}}{ 2}\right) - \sin x \times \left(\frac{1}{2}\right)\right)\\ y =\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x}-\sqrt{3}\cos x+\sin x ······\left(\mathrm{b}\right)\end{array}$

By comparing equation with equation, we find that $y=\sinh x-2 \cos \left(x+\frac{\pi }{6}\right)$is a particular solution for the given fourth order differential equation, where $k_1=\frac{1}{2}, k_2=-\frac{1}{2}, k_3=-\sqrt{3}\mathrm{and} k_4=1$.

$y=k_1{e}^x+k_2{e}^{-x}+k_3 \cos x+k_4\sin x$is the general solution. For$k_1=\frac{1}{2}, k_2=-\frac{1}{2}, k_3=-\sqrt{3}$and$k_4=1$in general value, we get the required particular solution.

Hence the solutions are

$\begin{array}{l}{k}_1=\frac{1}{2}, k_2=-\frac{1}{2}, k_3=-\sqrt{3} k_4=1\\ y =\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x}-\sqrt{3}\cos x+\sin x x\end{array}$