Question

`y'' + 9y = 2x^2 - 5`, How about solution. (`y_h`=?, `y_p=`?)

Answer 1

`y(x) = Acos(3x)+Bsin(3x) + 2/9x^2 -49/91`

Explanation:

We seek a solution to

`y'' + 9y = 2x^2 - 5` ..... [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 polynomial 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'' + 9y = 0` ..... [B]

And it's associated Auxiliary equation is:

`m^2+9 = 0`

And so we have the solutions:

`m = +--1 \ \ \ \` (pure imaginary)

The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.

• Real distinct roots `m=alpha,beta, ...` will yield linearly independent solutions of the form `y_1=Ae^(alphax)`, `y_2=Be^(betax)`, ...
• Real repeated roots `m=alpha`, will yield a solution of the form `y=(Ax+B)e^(alphax)` where the polynomial has the same degree as the repeat.
• Complex roots (which must occur as conjugate pairs) `m=p+-qi` will yield a pairs linearly independent solutions of the form `y=e^(px)(Acos(qx)+Bsin(qx))`

Thus the solution of the homogeneous equation [B] is:

`y = e^(0x)(Acos(3x)+Bsin(3x) )`
`\ \ = Acos(3x)+Bsin(3x)`

Particular Solution

In order to find a particular solution of the non-homogeneous equation:

`y'' + 9y = f(x) \ \` with `f(x) = 2x^2 - 5`

So, we should probably look for a solution of the form:

`y = ax^2+bx+c` ..... [C]

Where the constants `a,b,c,d` is to be determined by direct substitution and comparison:

Differentiating [C] wrt `x` twice we get:

`y^((1)) = 2ax+b`
`y^((2)) = 2a`

Substituting these results into the DE [A] we get:

`(2a) + 9(ax^2+bx+c) = 2x^2 - 5`

Equating coefficients we get:

`x^2: 9a = 2 => a = 2/9`
`x^1: 9b=0=> b =0`
`x^0: 2a+9c = -5 => c = -49/91`

And so we form the Particular solution:

`y_p = 2/9x^2 -49/91`

General Solution

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = Acos(3x)+Bsin(3x) + 2/9x^2 -49/91`

Note this solution has `2` constants of integration and `2` linearly independent solutions, hence by the Existence and Uniqueness Theorem their superposition is the General Solution