Answer to Question #146832 in Differential Equations for Nikhil

Question

Solve the differential equation:
(D^4-a^4)y=x^4+sinbx, where a and b are constants and D=d/dx

Answer 1

$$Solution$$

It's AE is $D^4-a^4=0$

$\therefore\ the\ auxiliary\ equation\ becomes\ (D^2 – a^2 ) (D^2 + a^2 ) = 0\\ \implies D^2-a^2=0\ or\ D^2+a^2=0\\ \therefore (D – a) (D + a) (D – ai) (D + ai) = 0\\ \therefore D = \pm a, and\ D=\pm ai$

Hence the required solution is

$C.F=[ae^{ax}+be^{-ax}]+[ce^{iax}+de^{-iax}]\\ C.F =[a \frac{(cosh\ ax+sinh\ ax)}{2}]+[b \frac{(cosh\ ax-sinh\ ax)}{2}]+[c \frac{(cos\ ax-i\ sin\ ax)}{2}+d \frac{(cos\ ax-sin\ ax)}{2}]\\ C.F=[\frac{a+b}{2}cos\ ax + \frac{a-b}{2}sinh\ ax]+[\frac{c+d}{2}cos\ ax + i\frac{(c-d)}{2}sin\ ax]$

Where $C_1=\frac{a+b}{2},\ C_2=\frac{a-b}{2},\ C_3=\frac{c+d}{2},\ c_4=\frac{i(c-d)}{2},\$

Its solution is

$$C.F=[C_1\ cosh\ ax +C_2\ sinh\ ax]+[C_3\ cos\ ax +C_4\ sin\ ax]---(i)$$

And PI, for $x^4$ is;

$$PI_1=\frac{1}{D^4-a^4}\ x^4$$

Putting $D^4=a^4$ in $f(D)$, then

$PI_1=\frac10 x^4\ \therefore method\ fails$

Then to find PI, put $D^4+a^4$ for $f(D)$ and take out $x^4$.

$\therefore PI_1=x^4 \{\frac{1}{D^4-a^4+a^4}\}\{1\}\\ \implies x^4\ \frac{1}{D^4}\{1\}=x^4\frac{x^4}{4!} \implies \frac{x^8}{4!}-----(ii)$

Also, $PI_2$ for $sin\ bx$ is;

$$\therefore PI_2=\frac{1}{D^4-a^4}\ sin\ bx\\$$

Put $D^4=D^2 \cdot D^2 = (-b^2)(-b^2)$

$\implies\frac{1}{(-b^2)(-b^2)-a^4}\ sin\ bx = \frac{1}{b^4-a^4}\ sin\ bx ---(iii)$

General Solution is;

$G.S=C.F+PI_1+PI_2 \implies [C_1\ cosh\ ax +C_2\ sinh\ ax]+[C_3\ cos\ ax +C_4\ sin\ ax]+\frac{x^8}{4!}+\frac{1}{b^4-a^4}\ sin\ bx$

Hence

$G.S=[C_1\ cosh\ ax +C_2\ sinh\ ax]+[C_3\ cos\ ax +C_4\ sin\ ax]+\frac{x^8}{4!}+\frac{1}{b^4-a^4}\ sin\ bx$