BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.1, Problem 62E
To determine

To find: The function y.

Expert Solution

Answer to Problem 62E

The function is y=12x2+12x+34.

Explanation of Solution

Given:

The differential equation is y+y2y=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)=nxn1

(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(2x21)+B(1x11)=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(1x11)=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+y2y=x2,

(2A)+(2Ax+B)2(Ax2+Bx+C)=x22A+2Ax+B2Ax22Bx2C=x22Ax2+(2A2B)x+(2A+B2C)=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.

2A2B=0

Substitute 12 for A,

2(12)2B=012B=02B=1B=12

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

2A+B2C=0

Substitute 12 for A. and 12 for B in 2A+B2C=0,

2(12)+(12)2C=0322C=02C=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=x22x234

Therefore, the required function is y=12x212x34.

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