Solve the following differential equation?

(D-1)^(2)[D^(2)+1)^(2)y=sin^(2)(x/2)+e^(x)+x

1 Answer
Jul 18, 2018

y = (Ax+B)sin(x) + (Cx+D)cos(x) + (Ex+F)e^x + 1/8x^2e^x - 1/32 x^2sin(x) + x + 5/2

for constants A, B, C, D, E, and F

Explanation:

We can rewrite this as
((D-1)(D^2+1))^2y = 1/2 - 1/2 cos(x) + e^x + x

Just looking at the left side,
(D^3 - D^2 + D - 1)^2y
= (D^6 - 2D^5 + 3D^4-4D^3+3D^2-2D + 1)y = Oy
(where we define O to be our operator for ease later)

Let's focus on the specific solution before we discuss homogenous solutions. It is logical to assume some functional forms of our solutions.

For the constant, clearly that constant can be our function, i.e.
O(a) = a \ \ forall a in mathbb(C) .

For a linear function, all higher order derivatives will be negligible, so we can try a form of y_0 = Ax + B. Plugging this in,
Oy_0 = -2A + Ax+B = Ax + (B-2A)

To get the constants above, therefore A = 1 and B = 5/2.

For an exponential function y_1 = Ce^x, D is an identity, hence
Oy_1 = (C - 2C + 3C-4C+3C-2C+C)e^x = 0

Therefore, we try y_2 = Fxe^x, which is a little less lovely until you realize that the nth derivative of xe^x equals xe^x + n e^x (try it!).

Therefore (the xe^x terms are identical to before and will therefore cancel)
Oy_2 = 6e^x - 10e^x + 12e^x - 12e^x + 6e^x - 2e^x = 0

Therefore, we will try y_3 = Gx^2e^x. Here, one can prove that
d^n/dx^n(x^2e^x) = x^2e^x +2nxe^x + n(n-1)e^x

Therefore, the first two terms will cancel out again, and
Oy_3 = G(30 - 2 * 20 + 3 * 12 - 4 * 6 + 3 * 2) = 8G
Therefore, G = 1/8. This need for x^2 is obvious at the factored version of the differential equation.

We can repeat this process for the sinusoid and we find that we need to use -1/32 x^2sin(x). Therefore, the specific solution is
y_c = 1/8x^2e^x - 1/32 x^2sin(x) + x + 5/2

As is apparent by the factored differential equation, we have eigenfunctions of sin(x), cos(x), e^x, and each of those multiplied by x.

Therefore, our general solution is
y = (Ax+B)sin(x) + (Cx+D)cos(x) + (Ex+F)e^x + y_c