Question

Solve
(D^4+2D^3-3D^2)y=x^2+3e^2x+4sinx

Answer 1

Auxiliary equation is $D^4 +2D^3-3D^2 = 0$ (1)

Solving it, we get $D^2(D-1)(D+3) = 0$

$D = 0,0,1,-3$

Then $y = c_1+c_2x+c_3e^{x}+c_4e^{-3x}$

For particular integral,

For $x^2$,

Let $y = ax^4 + bx^3+cx^2 +dx+e$ be it's solution

then

$y'' = 12ax^2 +6bx +2c$

$y''' = 24ax + 6b$

$y'''' = 24a$

Putting values in equation $D^4+2D^3-3D^2 = x^2$ , we get

$-36ax^2 + (48a-6b)x + (24a+12b-6c) = x^2$

comparing powers,

we get $a = -\frac{1}{36}, b = -\frac{2}{27}, c = -\frac{7}{27}$

then P.I. for $x^2$ is, $-\frac{1}{36}x^4 - \frac{2}{27}x^3 - \frac{7}{27}x^2$

P.I. for $3e^{2x}$ ,

$\frac{1}{D^4+2D^3-3D^2}e^{2x} = e^{2x}\frac{1}{2^4+(2\times 2^3)-(3\times 2^2)} = \frac{3}{20}e^{2x}$

P.I. for 4sin(x)

$\frac{1}{D^4+2D^3-3D^2}4sin(x) = \frac{1}{D^2(D^2+2D-3)}4sin(x)$

$= \frac{1}{(-1^2)(-1^2+2D-3)}4sin(x) = \frac{1}{2-D}2sin(x) = -2\frac{(D+2)}{D^2 - 4}sin(x)$

$= -2\frac{cos(x)+2sin(x)}{-1-4} = \frac{2}{5}(cos(x)+2sin(x))$

Solution of the differential equation is

$y = c_1+c_2x+c_3e^{x}+c_4e^{-3x}-\frac{1}{36}x^4 - \frac{2}{27}x^3 - \frac{7}{27}x^2+\frac{3}{20}e^{2x} +\frac{2}{5}(cos(x)+2sin(x))$