# Find the value of constant using given differential equation.

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 3.3, Problem 41E
To determine

Expert Solution

## Answer to Problem 41E

The value of A=310 and B=110 .

### Explanation of Solution

Given:

The given function is y=Asinx+Bcosx and the differential equation is y"+y'2y=sinx

Calculation:

y=Asinx+Bcosx

Apply sum rule.

(f+g)'=f'+g'

y'=Addx(sinx)+Bddx(cosx)

Use derivative rule ddx(sinx)=cosx and ddx(cosx)=sinx .

y'=Acosx+B(sinx)y'=AcosxBsinx

Apply difference rule.

(fg)'=f'g'

y''=AddxcosxBddxsinx

Use derivative rule ddx(sinx)=cosx and ddx(cosx)=sinx .

y''=A(sinx)Bcosxy''=AsinxBcosx

Now,

y"+y'2y=sinx

AsinxBcosx+AcosxBsinx2(Asinx+Bcosx)=sinxAsinxBcosx+AcosxBsinx2Asinx2Bcosx=sinx(AsinxBsinx2Asinx)+(Bcosx+Acosx2Bcosx)=sinx(3AB)sinx+(3B+A)cosx=1sinx+0cosx

Compare both sides.

3AB=1(i)A3B=0(ii)

Add equation (i) toequation (ii)×3 .

3AB=1[A3B=0]×3010B=11010B=110B=110

Substitute the value of B=110 in equation (i) .

3AB=13A(110)=13A+110=13A+110110=11103A=91033A=910×3A=310

Hence thevalue of A=310 and B=110 .

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