Question

Differential Equation

Answer 1

1. The given equation is probably supposed to read

y'' - 2y' - 3y = 64x exp(-x)

First consider the homogeneous equation,

y'' - 2y' - 3y = 0

which has characteristic equation

r² - 2r - 3 = (r - 3) (r + 1) = 0

with roots r = 3 and r = -1. Then the characteristic solution is

$y = C_1 e^{3x} + C_2 e^{-x}$

and we let y₁ = exp(3x) and y₂ = exp(-x), our fundamental solutions.

Now we use variation of parameters, which gives a particular solution of the form

$y_p = u_1y_1 + u_2y_2$

where

$\displaystyle u_1 = -\int \frac{64xe^{-x}y_2}{W(y_1,y_2)} \, dx$

$\displaystyle u_2 = \int \frac{64xe^{-x}y_1}{W(y_1,y_2)} \, dx$

and W(y₁, y₂) is the Wronskian determinant of the two fundamental solutions. This is

$W(y_1,y_2) = \begin{vmatrix}e^{3x} & e^{-x} \\ (e^{3x})' & (e^{-x})'\end{vmatrix} = \begin{vmatrix}e^{3x} & e^{-x} \\ 3e^{3x} & -e^{-x}\end{vmatrix} = -e^{2x} - 3e^{2x} = -4e^{2x}$

Then we find

$\displaystyle u_1 = -\int \frac{64xe^{-x} \cdot e^{-x}}{-4e^{2x}} \, dx = 16 \int xe^{-4x} \, dx = -(4x + 1) e^{-4x}$

$\displaystyle u_2 = \int \frac{64xe^{-x} \cdot e^{3x}}{-4e^{2x}} \, dx = -16 \int x \, dx = -8x^2$

so it follows that the particular solution is

$y_p = -(4x+1)e^{-4x} \cdot e^{3x} - 8x^2\cdot e^{-x} = -(8x^2+4x+1)e^{-x}$

and so the general solution is

$\boxed{y(x) = C_1 e^{3x} + C_2e^{-x} - (8x^2+4x+1) e^{-x}}$

2. I'll again assume there's typo in the equation, and that it should read

y''' - 6y'' + 11y' - 6y = 2x exp(-x)

Again, we consider the homogeneous equation,

y''' - 6y'' + 11y' - 6y = 0

and observe that the characteristic polynomial,

r³ - 6r² + 11r - 6

has coefficients that sum to 1 - 6 + 11 - 6 = 0, which immediately tells us that r = 1 is a root. Polynomial division and subsequent factoring yields

r³ - 6r² + 11r - 6 = (r - 1) (r² - 5r + 6) = (r - 1) (r - 2) (r - 3)

and from this we see the characteristic solution is

$y_c = C_1 e^x + C_2 e^{2x} + C_3 e^{3x}$

For the particular solution, I'll use undetermined coefficients. We look for a solution of the form

$y_p = (ax+b)e^{-x}$

whose first three derivatives are

${y_p}' = ae^{-x} - (ax+b)e^{-x} = (-ax+a-b)e^{-x}$

${y_p}'' = -ae^{-x} - (-ax+a-b)e^{-x} = (ax-2a+b)e^{-x}$

${y_p}''' = ae^{-x} - (ax-2a+b)e^{-x} = (-ax+3a-b)e^{-x}$

Substituting these into the equation gives

$(-ax+3a-b)e^{-x} - 6(ax-2a+b)e^{-x} + 11(-ax+a-b)e^{-x} - 6(ax+b)e^{-x} = 2xe^{-x}$

$(-ax+3a-b) - 6(ax-2a+b) + 11(-ax+a-b) - 6(ax+b) = 2x$

$-24ax+26a-24b = 2x$

It follows that -24a = 2 and 26a - 24b = 0, so that a = -1/12 = -12/144 and b = -13/144, so the particular solution is

$y_p = -\dfrac{12x+13}{144}e^{-x}$

and the general solution is

$\boxed{y = C_1 e^x + C_2 e^{2x} + C_3 e^{3x} - \dfrac{12x+13}{144} e^{-x}}$