Question

Write the general solution of the fourth-order DE $y^{\left(4\right)}-2y"+y=0$ entirely in terms of hyperbolic functions.

(b) Write down the form of a particular solution of$y^{\left(4\right)}-2y"+y=sinhx$.

Short answer

The particular solution is$y_p=A{x}^2\cos hx+B{x}^2\sin hx$

Step-by-step solution

Define the particular solution

A Particular Solution is a differential equation solution that is derived from the General Solution by assigning specific values to the random constants. Depending on the query, the requirements for finding the values of the random constants can be submitted to us as an Initial-Value Problem or Boundary Conditions.

Now find the general solution

(a) Assume that $y=e^{mx}$and then differentiate with respect to x as

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

Substitute with our assumption $y=e^{mx}$

$$\begin{gathered} m^4{e}^{mx}-2m^2{e}^{mx}+e^{mx}=0 \\ \left(m^2-2m^2+1\right)e^{mx}=0 \end{gathered}$$

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

$$\begin{gathered} m^4-2m^2+1=0\left(m^2-1\right)\left(m^2-1\right)=0 \\ \left(m-1\right)\left(m+1\right)(m-1)(m+1)=0 \\ {\left(m-1\right)}^2{\left(m\_1\right)}^2=0 \end{gathered}$$

Then we have the roots

role="math" localid="1667906668854" $m_{1,2}=1and{m}_{3,4}=-1$

Then we can obtain the general solution of the given differential equation as

$y=c_1{e}^x+c_{2}x{e}^x+c_3{e}^{-x}+c_{4}x{e}^{-x}$

Now find the particular solution

We have

$$\begin{gathered} c_1=\left(\frac{k_1}{2}+\frac{k_2}{2}\right)\text{'} \\ {c}_3=\left(\frac{k_1}{2}-\frac{k_2}{2}\right) \\ {c}_2=\left(\frac{k_3}{2}+\frac{k_4}{2}\right) \\ {c}_4=\left(\frac{k_1}{2}-\frac{k_2}{2}\right) \\ y=\left(\frac{k_1}{2}+\frac{k_2}{2}\right)e^x+\left(\frac{k_3}{2}+\frac{k_4}{2}\right)x{e}^x+\left(\frac{k_1}{2}-\frac{k_2}{2}\right)e^{-x}+\left(\frac{k_1}{2}-\frac{k_2}{2}\right)x{e}^{-x} \\ =\frac{k_1}{2}\left(e^x+e^{-x}\right)+\frac{k_2}{2}\left(e^x-e^{-x}\right)+\frac{k_3}{2}x\left(e^x+e^{-x}\right)+\frac{k_4}{2}x\left(e^x-e^{-x}\right) \\ =k_1\left(\frac{e^x+e^{-x}}{2}\right)+k_2\left(\frac{e^x-e^{-x}}{2}\right)+k_{3}x\left(\frac{e^x+e^{-x}}{2}\right)+k_{4}x\left(\frac{e^x-e^{-x}}{2}\right) \\ =k_1\cos hx+k_2\sin hx+k_{3}x\cos hx+k_{4}x\sin hx \end{gathered}$$

b) If we have the non-homogeneous differential equation$y"-2y\text{'}+y=\sin hx$,

then we can write its particular solution as

$y_p=A{x}^2\cos hx+B{x}^2\sin hx$