Question

The equation \(y'' + y' - 2y = {x^2}\) 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 \(y = A{x^2} + Bx + C\) satisfies this equation. (Differential equations will be studied in detail in Chapter 9.).

Short answer

The values of constants are:

\(\begin{array}{l}A &=& - \frac{1}{2}\\B &=& - \frac{1}{2}\\C &=& - \frac{3}{4}\end{array}\)v

Step-by-step solution

Differential Equation

The differential equation is a form of the algebraic equation that consists of functions formed using the derivative of the functions connected with mathematical operation between them.

Solving for constants.

Thegiven function is \(y = A{x^2} + Bx + C\).

Differentiating with respect to\(x\)as:

\(\begin{align}y'\left( x \right) &= 2Ax + B\\y''\left( x \right) &= 2A\end{align}\)

Substituting these values in the equation \(y'' + y' - 2y = {x^2}\), we have:

\(\begin{align}2A + 2Ax + B - 2A{x^2} - 2Bx - 2C &= {x^2}\\\left( { - 2A} \right){x^2} + \left( {2A - 2B} \right)x + \left( {2A + B - 2C} \right) &= {x^2} + 0x + 0\end{align}\)

Comparing the obtained equation.

Comparing both sides, we have:

\(\begin{array}{c}\left( { - 2A} \right) &=& 1 \Rightarrow A &=& - \frac{1}{2}\\\left( {2A - 2B} \right) &=& 0 \Rightarrow B &=& - \frac{1}{2}\\\left( {2A + B - 2C} \right) &=& 0 \Rightarrow C &=& - \frac{3}{4}\end{array}\)

So, the constants are:

\(\begin{array}{l}A &=& - \frac{1}{2}\\B &=& - \frac{1}{2}\\C &=& - \frac{3}{4}\end{array}\)

Hence, this is the required answer.