#### To determine

**To find:** The function *y*.

#### Answer

The function is y=−12x2+−12x+−34.

#### Explanation

**Given:**

The differential equation is y″+y′−2y=x2 and the function y=Ax2+Bx+C, where *A*, *B*, and *C* are constant.

**Derivative rules:**

(1) Constant multiple rule: ddx(cf)=cddx(f)

(2) Power rule: ddx(xn)=nxn−1

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

**Calculation:**

Obtain the first derivative of y=Ax2+Bx+C.

y′(x)=ddx(Ax2+Bx+C)

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

y′(x)=ddx(Ax2)+ddx(Bx)+ddx(C)=Addx(x2)+Bddx(x)+ddx(C)

Since the derivative of constant function is zero, then ddx(C)=0.

y′(x)=Addx(x2)+Bddx(x)+0=Addx(x2)+Bddx(x)

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

y′(x)=A(2x2−1)+B(1x1−1)=A(2x)+B(1)=2Ax+B

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

Obtain the second derivative of y=Ax2+Bx+C.

y″(x)=ddx(dydx)=ddx(2Ax+B)

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

y″(x)=ddx(2Ax)+ddx(C)=2Addx(x)+ddx(C)

Since derivative of constant function is zero, then ddx(C)=0.

y″(x)=2Addx(x)+0

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

y″(x)=2A(1x1−1)=2A(1)=2A

Therefore, the second derivative of y is 2A.

Obtain the required function.

Substitute the values of y(x), y′(x) and y″(x) in y″+y′−2y=x2,

(2A)+(2Ax+B)−2(Ax2+Bx+C)=x22A+2Ax+B−2Ax2−2Bx−2C=x2−2Ax2+(2A−2B)x+(2A+B−2C)=x2

The left hand side coefficient of x2 is equal to the right hand side coefficient of x2.

−2A=1A=−12

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

2A−2B=0

Substitute −12 for *A*,

2(−12)−2B=0−1−2B=0−2B=1B=−12

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

2A+B−2C=0

Substitute −12 for *A*. and −12 for *B* in 2A+B−2C=0,

2(−12)+(−12)−2C=0−32−2C=0−2C=32C=−34

Therefore, the value of *A*, *B*, and *C* are −12,−12, and −34.

Substitute the value of *A*, *B*, *C* in y=Ax2+Bx+C,

y=−12x2+−12x+−34=−x22−x2−34

Therefore, the required function is y=−12x2−12x−34.