Question

Solve the linear partial differential equation (D⁴+D'⁴)=0

Answer 1

Solution;

Given;

$D^4+D'^4=0$

The auxiliary equation is;

$m^4+1=0$

$m^4=-1=e^{iπ}$ (By Euler's formula)

By De Moivres theorem;

$m^4=e^{iπ+2nπ}$

$m=e^{iπ(\frac{2n+1}{4})}$ In which n=0,1,2,3....

When;

n=0;

$m=e^{i\fracπ4}=cos\fracπ4+isin\fracπ4=\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}$

n=1;

$m=e^{i(\frac{3π}{4})}=cos\frac{3π}{4}+isin\frac{3π}{4}=\frac{-1}{\sqrt2}+i\frac{1}{\sqrt2}$

n=2;

$m=e^{i(\frac{5π}{4})}=cos\frac{5π}{4}+i sin\frac{5π}{4}=\frac{-1}{\sqrt2}-i\frac{1}{\sqrt2}$

n=3;

$m=e^{i (\frac{7π}{4})}=cos\frac{7π}{4}+i sin\frac{7π}{4}=\frac{1}{\sqrt2}-i\frac{1}{\sqrt2}$

Therefore,you can see that the roots are;

$m=\frac{1}{\sqrt2}_-^+i\frac{1}{\sqrt2}$ and $-\frac{1}{\sqrt2}_-^+\frac{1}{\sqrt2}$

Therefore,the complementary solution is;

$C.F=e^{\frac{1}{\sqrt2}x}[c_1cos\frac{1}{\sqrt2}x+c_2sin\frac{1}{\sqrt2}x]+e^{\frac{-1}{\sqrt2}}[c_3cos\frac{1}{\sqrt2}x+c_4sin\frac{1}{\sqrt2}x]$