Q30E. - Fundamentals Of Differential Equations And Boundary Value Problems

Question

(a) Derive the form  $y\left(x\right)=A_1{e}^x+A_2{e}^{-x}+A_{3}cosx+A_{4}sinx$ for the general solution to the equation  $y^{\left(4\right)}=y$, from the observation that the fourth roots of unity are 1, -1, i, and -i. 

 

(b) Derive the form 

 $y\left(x\right)=A_1{e}^x+A_2{e}^{-x/2}cos\left(\frac{\sqrt{3x}}{2}\right)+A_3{e}^{-x/2}sin\left(\frac{\sqrt{3x}}{2}\right)$

for the general solution to the equation  $y^{\left(3\right)}=y$ from the observation that the cube roots of unity are 1, $e^{i\frac{2\pi }{3}}$  , and  $e^{-i\frac{2\pi }{3}}$.

Step-by-Step Solution

Answer

Proved.

Step 1: Solving for (a):

 

(a)Let differential equation be,

 $y^{\left(4\right)}=y$

Then its corresponding auxillary equation is,

 $\begin{array}{l}{m}^4=1\\ m=1,-1,i,-i\end{array}$

 

Let,

 $\begin{array}{l}{m}_1=1\\ m_2=-1\\ m_3=i\\ m_4=-i\end{array}$

 

Here, two roots are real and distinct and other two roots are purely complex and distinct. 

Then the general solution is given by,

 

           $y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}+c_3{e}^{m_{3}x}+c_4{e}^{m_{4}x}$             ( $c_1,c_2,c_3,c_4$ are constant coefficient)

$\begin{array}{l}=c_1{e}^x+c_2{e}^{-x}+c_3{e}^{ix}+c_4{e}^{-ix}\\ =c_1{e}^x+c_2{e}^{-x}+c_3\left(cos\right(x)+isin(x\left)\right)+c_4\left(cos\right(-x)+isin(-x\left)\right)\\ =c_1{e}^x+c_2{e}^{-x}+cos\left(x\right)(c_3+c_4)+sin\left(x\right)(c_{3}i+c_{4}i)\end{array}$

Let $A_3=c_3+c_4$   and  $A_4=c_{3}i+c_{4}i$

 

Then,

 $y=c_1{e}^x+c_2{e}^{-x}+A_3\cos \left(x\right)+A_4\sin \left(x\right)$

Step 2: proving further:

Let   $c_1=A_1$ and   $c_2=A_2$

 $y=A_1{e}^x+A_2{e}^{-x}+A_3\cos \left(x\right)+A_4\sin \left(x\right)$

Hence proved

Step 3: Solving for (b):

(b)Let the given differential equation be,

 $y^{\left(3\right)}=y$

Then its corresponding auxillary equation is,

 $\begin{array}{l}{m}^3=1\\ y\text{'}=1,e^{\frac{2\pi i}{3}},e^{-\frac{2\pi i}{3}}\end{array}$

 

Let,

 $\begin{array}{l}{m}_1=1\\ m_2=e^{\frac{2\pi i}{3}}\\ m_3=e^{-\frac{2\pi i}{3}}\end{array}$

 

Here, two roots are real and distinct and other two roots are purely complex and distinct.

Therefore, the general solution is given by

 $y=A_1{e}^{m_{1}x}+C{e}^{m_{2}x}+D{e}^{m_{3}x}$

Here now to find exact value of  $e^{\frac{2\pi i}{3}}$ and  $e^{-\frac{2\pi i}{3}}$

 $\begin{array}{l}{e}^{\frac{2\pi i}{3}}=\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\\ =\cos (\pi -\frac{\pi }{3})+i\sin (\pi -\frac{\pi }{3})\\ =\cos \left(\pi \right)\cos \left(\frac{\pi }{3}\right)+\sin \left(\pi \right)\sin \left(\frac{\pi }{3}\right)+i(\sin \left(\pi \right)\cos \left(\frac{\pi }{3}\right)-\cos \left(\pi \right)\sin \left(\frac{\pi }{3}\right))\\ =(-1.\frac{1}{2}+0)+i(0-\frac{\sqrt{3}}{2}.-1)\\ e^{\frac{2\pi i}{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}\end{array}$

 

On similar lines

$e^{\frac{2\pi i}{3}}=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$ …(3)

Step 3: Solving by substituting further:

Substituting (2) and (3) in (1) we get,

$\begin{array}{l}y=A_1{e}^x+C{e}^{(-\frac{1}{2}+i\frac{\sqrt{3}x}{2})}+D{e}^{(-\frac{1}{2}-i\frac{\sqrt{3}x}{2})}\\ =A_1{e}^x+C[e^{-\frac{1}{2}x}.e^{i\frac{\sqrt{3}x}{2}}]+D[e^{-\frac{1}{2}x}.e^{-i\frac{\sqrt{3}x}{2}}]\\ =A_1{e}^x+C{e}^{-\frac{1}{2}x}(\cos \left(\frac{\sqrt{3}x}{2}\right)+i\sin \left(\frac{\sqrt{3}x}{2}\right))+D{e}^{-\frac{1}{2}x}(\cos (-\frac{\sqrt{3}x}{2})+i\sin (-\frac{\sqrt{3}x}{2}))\\ =A_1{e}^x+C{e}^{-\frac{1}{2}x}(\cos \left(\frac{\sqrt{3}x}{2}\right)+i\sin \left(\frac{\sqrt{3x}}{2}\right))+D{e}^{-\frac{1}{2}x}(\cos \left(\frac{\sqrt{3x}}{2}\right)-i\sin \left(\frac{\sqrt{3}x}{2}\right))\\ =A_1{e}^x+e^{-\frac{1}{2}x}[C(\cos \left(\frac{\sqrt{3}x}{2}\right)+i\sin \left(\frac{\sqrt{3}x}{2}\right))+D(\cos \left(\frac{\sqrt{3}x}{2}\right)-i\sin \left(\frac{\sqrt{3}x}{2}\right))]\\ =A_1{e}^x+e^{-\frac{1}{2}x}[\cos \left(\frac{\sqrt{3}x}{2}\right)(Ci+Di)+\sin \left(\frac{\sqrt{3}x}{2}\right)(Ci-Di)]\end{array}$

Let  $A_2=C+D$ and  $A_3=Ci-Di$

 $\begin{array}{l}\Rightarrow y=A_1{e}^x+e^{-\frac{1}{2}x}[\cos \left(\frac{\sqrt{3}x}{2}\right)A_2+\sin \left(\frac{\sqrt{3}x}{2}\right)A_3]\\ =A_1{e}^x+A_2{e}^{-\frac{1}{2}x}\cos \left(\frac{\sqrt{3}x}{2}\right)+A_3{e}^{-\frac{1}{2}x}\sin \left(\frac{\sqrt{3}x}{2}\right)\end{array}$

$y=A_1{e}^x+A_2{e}^{-\frac{1}{2}x}\cos \left(\frac{\sqrt{3}x}{2}\right)+A_3{e}^{-\frac{1}{2}x}\sin \left(\frac{\sqrt{3}x}{2}\right)$

 

Hence proved.