# Suppose that x^2 + \cos (x^2) is a solution to the DE y'' - 8y' + 16 = h(x). (a) Find the value...

## Question:

Suppose that {eq}\displaystyle x^2 + \cos (x^2){/eq} is a solution to the {eq}\displaystyle DE y'' - 8y' + 16 = h(x){/eq}. (a) Find the value of h(x). (b) Find the general solution to the DE.

## Linear Inhomogeneous Differential Equations:

An {eq}n {/eq}th-order linear inhomogeneous differential equation for the unknown function {eq}y {/eq} is an equation of the form

{eq}\displaystyle \sum_{k=0}^n p_k(x)\frac{d^ky}{dx^k}=q(x) {/eq}

where {eq}p_0,p_1,\dots,p_n,q {/eq} are all continuous functions, and neither {eq}p_n(x) {/eq} nor {eq}q(x) {/eq} is the zero function. The function {eq}q(x) {/eq} is called the inhomogeneous term of this equation, while the equation

{eq}\displaystyle \sum_{k=0}^n p_k(x)\frac{d^ky}{dx^k}=0 {/eq}

is called the associated homogeneous equation.

If {eq}y_p {/eq} is any one solution to a linear inhomogeneous differential equation, then every solution to that equation can be written in the form {eq}y=y_p+y_h {/eq}, where {eq}y_h {/eq} is some solution to the associated homogeneous equation.

## Answer and Explanation:

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(a) If {eq}y=x^2+\cos(x^2) {/eq}, then differentiating {eq}y {/eq} twice gives:

{eq}\begin{align*} y'&=2x-2x\sin(x^2)\\ y''&=2-2\sin(x^2)-2x(2x\c...

See full answer below.

First-Order Linear Differential Equations

from

Chapter 16 / Lesson 3
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In this lesson you'll learn how to solve a first-order linear differential equation. We first define what such an equation is, and then we give the algorithm for solving one of that form. Specific examples follow the more general description of the method.