Chapter 3.1, Problem 62E

The equation y" + y' – 2y = x² is called a differential equation because it involves an unknown function y and its derivatives y' and y". Find constants A, B. and C such that the function y = Ax² + Bx + C satisfies this equation. (Differential equations will he studied in detail in Chapter 9.)

To determine

To find: The function y.

Answer to Problem 62E

The function is $y=\frac{-1}{2}{x}^2+\frac{-1}{2}x+\frac{-3}{4}$.

Explanation of Solution

Given:

The differential equation is $y^{″}+y^{\prime }-2y=x^2$ and the function $y=A{x}^2+Bx+C$, where A, B, and C are constant.

Derivative rules:

(1) Constant multiple rule: $\frac{d}{dx}\left(cf\right)=c\frac{d}{dx}\left(f\right)$

(2) Power rule: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

(3) Sum rule: $\frac{d}{dx}\left(f+g\right)=\frac{d}{dx}\left(f\right)+\frac{d}{dx}\left(g\right)$

Calculation:

Obtain the first derivative of $y=A{x}^2+Bx+C$.

$y^{\prime }{\left(x\right)}^{}=\frac{d}{dx}\left(A{x}^2+Bx+C\right)$

Apply the sum rule (3) and the constant multiple rule (1),

$\begin{array}{c}{y}^{\prime }\left(x\right)=\frac{d}{dx}\left(A{x}^2\right)+\frac{d}{dx}\left(Bx\right)+\frac{d}{dx}\left(C\right)\\ =A\frac{d}{dx}\left(x^2\right)+B\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(C\right)\end{array}$

Since the derivative of constant function is zero, then $\frac{d}{dx}\left(C\right)=0$.

$\begin{array}{c}{y}^{\prime }\left(x\right)=A\frac{d}{dx}\left(x^2\right)+B\frac{d}{dx}\left(x\right)+0\\ =A\frac{d}{dx}\left(x^2\right)+B\frac{d}{dx}\left(x\right)\end{array}$

Apply the power rule (2) and simplify the terms,

$\begin{array}{c}{y}^{\prime }{\left(x\right)}^{}=A\left(2x^{2-1}\right)+B\left(1x^{1-1}\right)\\ =A\left(2x\right)+B\left(1\right)\\ =2Ax+B\end{array}$

Therefore, the first derivative of $y$ is $2Ax+B$.

Obtain the second derivative of $y=A{x}^2+Bx+C$.

$\begin{array}{c}{y}^{″}\left(x\right)=\frac{d}{dx}\left(\frac{dy}{dx}\right)\\ =\frac{d}{dx}\left(2Ax+B\right)\end{array}$

Apply the sum rule (3) and the constant multiple rule (1),

$\begin{array}{c}{y}^{″}\left(x\right)=\frac{d}{dx}\left(2Ax\right)+\frac{d}{dx}\left(C\right)\\ =2A\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(C\right)\end{array}$

Since derivative of constant function is zero, then $\frac{d}{dx}\left(C\right)=0$.

$y^{″}\left(x\right)=2A\frac{d}{dx}\left(x\right)+0$

Apply the power rule (2) and simplify the terms,

$\begin{array}{c}{y}^{″}\left(x\right)=2A\left(1x^{1-1}\right)\\ =2A\left(1\right)\\ =2A\end{array}$

Therefore, the second derivative of $y$ is $2A$.

Obtain the required function.

Substitute the values of $y\left(x\right)$, $y^{\prime }\left(x\right)$ and $y^{″}\left(x\right)$ in $y^{″}+y^{\prime }-2y=x^2$,

$\begin{array}{l}\left(2A\right)+\left(2Ax+B\right)-2\left(A{x}^2+Bx+C\right)=x^2\\ 2A+2Ax+B-2A{x}^2-2Bx-2C=x^2\\ -2A{x}^2+\left(2A-2B\right)x+\left(2A+B-2C\right)=x^2\end{array}$

The left hand side coefficient of $x^2$ is equal to the right hand side coefficient of $x^2$.

$\begin{array}{l}-2A=1\\ A=-\frac{1}{2}\end{array}$

The left hand side coefficient of x is equal to right hand side coefficient of x.

$2A-2B=0$

Substitute $\frac{-1}{2}$ for A,

$\begin{array}{l}2\left(\frac{-1}{2}\right)-2B=0\\ -1-2B=0\\ -2B=1\\ B=\frac{-1}{2}\end{array}$

The left hand side of constant term is equal to the right hand side of constant term.

$2A+B-2C=0$

Substitute $\frac{-1}{2}$ for A. and $\frac{-1}{2}$ for B in $2A+B-2C=0$,

$\begin{array}{l}2\left(\frac{-1}{2}\right)+\left(\frac{-1}{2}\right)-2C=0\\ \frac{-3}{2}-2C=0\\ -2C=\frac{3}{2}\\ C=-\frac{3}{4}\end{array}$

Therefore, the value of A, B, and C are $\frac{-1}{2}$,$\frac{-1}{2}$, and $\frac{-3}{4}$.

Substitute the value of A, B, C in $y=A{x}^2+Bx+C$,

$\begin{array}{c}y=\frac{-1}{2}{x}^2+\frac{-1}{2}x+\frac{-3}{4}\\ =-\frac{x^2}{2}-\frac{x}{2}-\frac{3}{4}\end{array}$

Therefore, the required function is $y=-\frac{1}{2}{x}^2-\frac{1}{2}x-\frac{3}{4}$.