Question

Solve the differential equation `(D^2+2D+5)y=xe^x` ?

Answer 1

`y = Ae^(-x)cos2x + Be^(-x)sin2x + (xe^x)/8 - e^x/16`

Explanation:

We have:

`(D^2+2D+5)y=xe^x`

Where `D` is the linear differential operator `d/dx`. Thus we can write the equation as:

`(d^2y)/(dx^2) + 2dy/dx + 5y = xe^x` ..... [A]

This is a second order linear non-Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution, `y_c` of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, `y_p` of the non-homogeneous equation.

Complimentary Function

The homogeneous equation associated with [A] is

`(d^2y)/(dx^2) + 2dy/dx + 5y = 0`

And it's associated Auxiliary equation is:

`m^2 + 2m+5 = 0`

Which has two complex solutions `m=-1+-2i`

Thus the solution of the homogeneous equation is:

`y_c = e^(-1x)(Acos2x + Bsin2x)`
`\ \ \ = Ae^(-x)cos2x + Be^(-x)sin2x`

Particular Solution

In order to find a particular solution of the non-homogeneous equation we would look for a solution of the form:

`y = (ax+b)e^x`

Where the constants `a` and `b` are to be determined by direct substitution and comparison:

Differentiating wrt `x` (using the product rule) we get:

`dy/dx = (ax+b)e^x+(a)e^x`
`" " = (ax+a+b)e^x`

Differentiating again wrt `x` (using the product rule) we get:

`(d^2y)/(dx^2) = (ax+a+b)e^x+(a)e^x`
`" " = (ax+2a+b)e^x`

Substituting into the DE [A] we get:

`(ax+2a+b)e^x + 2(ax+a+b)e^x +5(ax+b)e^x = xe^x`

`:. (ax+2a+b) + 2(ax+a+b) +5(ax+b) = x`

Equating coefficients of `x` and constants we get

`x^1: (2a+b) + 2(a+b) +5(b) = 0 => 4a+8b=0`
`x^0: (a) + 2(a) +5(a) = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \=> 8a=1`

Solving simultaneously, we have:

`a=1/8`
`1/2+8b=0=>b=-1/16`

And so we form the Particular solution:

`y_p = (1/8x-1/6)e^x`
`\ \ \ = (xe^x)/8 - e^x/16`

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = Ae^(-x)cos2x + Be^(-x)sin2x + (xe^x)/8 - e^x/16`