What’s the solution of the given differential equation by method of variation of a parameter d2ydx2+3dydx+2y=e−xx?

Answer 1

Given the second order linear differential equation above...

∂²y/∂x² + 3∂y/∂x + 2y = xe^-x

We see that there are two solutions for y ( the homogeneous and the particular ). First let's solve the homogeneous solution (which is equivalent to solving the differential equation equal to zero).

∂²y/∂x² + 3∂y/∂x + 2y = 0

Using the ansatz of y(x) = e^rx due to the constant coefficients in the differential equation...

r²e^rx + 3re^rx + 2e^rx = 0

Since erx can never be zero (except at x=infinity) we have the simple quadratic equation to solve for r...

r² + 3r + 2 = 0 --> (r+1)(r+2)

r = -1, -2

y₁(x) = e^-x, y₂(x) = e^-2x

These represent the potential homogeneous solution which can be written as

y_H(x) = Ae^-x + Be^-2x

No we deal with the forcing function (xe^-x) to see if we can find a representation via a educated guess for the particular solution...

let y_P(x) = Cxe^-x

y'(x) = Ce^-x - Cxe^-x

y''(x) = -Ce^-x -Ce^-x + Cxe^-x = -2Ce^-x + Cxe^-x

Substitute into the differential equation to see if a value of C can be uniquely found...

-2Ce^-x + Cxe^-x + 3 * ( Ce^-x - Cxe^-x ) + 2Cxe^-x = xe^-x

Ce^-x + 0 ≠ xe^-x

Therefore we cannot find a C that fits this condition and hence our guess is too restrictive...

Instead lets try y_P(x) = Cx²e^-x + Dxe^-x

y'(x) = 2Cxe^-x - Cx²e^-x + De^-x - Dxe-x = Cx²e^-x + (2C - D)xe^-x + De^-x

y''(x) = 2Cxe^-x - Cx²e^-x + (2C - D)e^-x - (2C - D)xe^-x - De^-x

Placing these derivatives into the equation with the forcing function yields...

C = 1/2, D = -1

Therefore the final solution = y_H(x) + y_P(x) = Ae^-x + Be^-2x + (1/2)x²e^-x - xe^-x