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

May 17, 2018

$y \left(x\right) = A \cos \left(3 x\right) + B \sin \left(3 x\right) + \frac{2}{9} {x}^{2} - \frac{49}{91}$

#### Explanation:

We seek a solution to

$y ' ' + 9 y = 2 {x}^{2} - 5$ ..... [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

$y ' ' + 9 y = 0$ ..... [B]

And it's associated Auxiliary equation is:

${m}^{2} + 9 = 0$

And so we have the solutions:

$m = \pm - 1 \setminus \setminus \setminus \setminus$ (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 $m = \alpha , \beta , \ldots$ will yield linearly independent solutions of the form ${y}_{1} = A {e}^{\alpha x}$, ${y}_{2} = B {e}^{\beta x}$, ...
• Real repeated roots $m = \alpha$, will yield a solution of the form $y = \left(A x + B\right) {e}^{\alpha x}$ where the polynomial has the same degree as the repeat.
• Complex roots (which must occur as conjugate pairs) $m = p \pm q i$ will yield a pairs linearly independent solutions of the form $y = {e}^{p x} \left(A \cos \left(q x\right) + B \sin \left(q x\right)\right)$

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

$y = {e}^{0 x} \left(A \cos \left(3 x\right) + B \sin \left(3 x\right)\right)$
$\setminus \setminus = A \cos \left(3 x\right) + B \sin \left(3 x\right)$

Particular Solution

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

$y ' ' + 9 y = f \left(x\right) \setminus \setminus$ with $f \left(x\right) = 2 {x}^{2} - 5$

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

$y = a {x}^{2} + b x + c$ ..... [C]

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

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

${y}^{\left(1\right)} = 2 a x + b$
${y}^{\left(2\right)} = 2 a$

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

$\left(2 a\right) + 9 \left(a {x}^{2} + b x + c\right) = 2 {x}^{2} - 5$

Equating coefficients we get:

${x}^{2} : 9 a = 2 \implies a = \frac{2}{9}$
${x}^{1} : 9 b = 0 \implies b = 0$
${x}^{0} : 2 a + 9 c = - 5 \implies c = - \frac{49}{91}$

And so we form the Particular solution:

${y}_{p} = \frac{2}{9} {x}^{2} - \frac{49}{91}$

General Solution

Which then leads to the GS of [A}

$y \left(x\right) = {y}_{c} + {y}_{p}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = A \cos \left(3 x\right) + B \sin \left(3 x\right) + \frac{2}{9} {x}^{2} - \frac{49}{91}$

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