Question

What is the solution to the Differential Equation `(d^2y)/(dx^2) - 6dy/dx = 54x + 18`?

Answer 1

`y(x) = A + Be^(6x)-9/2x^2-9/2x`

Explanation:

We have:

`(d^2y)/(dx^2) - 6dy/dx = 54x + 18`

This is a second order linear non-Homogeneous Differentiation Equation. 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.

Complementary Function

The homogeneous equation associated with [A] is

`y''-6y'+0y = 0`

And it's associated Auxiliary equation is:

`m^2-6m+0 = 0`
`m^2-6m = 0`
`m(m-6) = 0`

Which has two real and distinct solutions `m0,6`

Thus the solution of the homogeneous equation is:

`y_c = Ae^(0x) + Be^(6x)`
`\ \ \ = A + Be^(6x)`

Particular Solution

With this particular equation [A], a probable solution is of the form:

`y = ax^2+bx +c`

Where `a,b,c` are constants to be determined by substitution

Let us assume the above solution works, in which case be differentiating wrt `x` we have:

`y' \ \= 2ax+b`
`y'' = 2a`

Substituting into the initial Differential Equation `[A]` we get:

`2a - 6(2ax+b) = 54x + 18`
`:. 2a - 12ax-6b = 54x + 18`

Equating coefficients of `x^0` and `x` we get:

`x^0: 2a-6b=18`
`x^1: -12a=54`

Solving simultaneous we have:

`a = -9/2, b = -9/2`

And so we form the Particular solution:

`y_p = -9/2x^2-9/2x`

General Solution

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = A + Be^(6x)-9/2x^2-9/2x`

As we have a linear combination of three linearly independent solutions, this is the GS.