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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Solve the differential equation or initial-value problem using the method of undetermined coefficients.

10. y" + y' – 2y = x + sin 2x, y(0) = 1, y'(0) = 0

To determine

To solve: The differential equation or initial-value problem by the method of undetermined coefficients.

Explanation

Given data:

The differential equation is,

y+y2y=x+sin2x (1)

with y(0)=1andy(0)=0 .

Consider the auxiliary equation is,

r2+r2=0 (2)

Roots of equation (2) are,

r=1±(1)24(1)(2)2(1){r=b±b24ac2afortheequationofar2+br+c=0}=1±32=1and2

Write the expression for the complementary solution,

yc(x)=c1er1x+c2er2x

Substitute 1 for r1 and 2 for r2 ,

yc(x)=c1ex+c2e2x

The particular solution yp1(x) is,

yp1(x)=Ax+B (3)

Differentiate equation (3) with respect to x,

yp1(x)=ddx(Ax+B)

yp1(x)=A (4)

Differentiate equation (4) with respect to x,

yp1(x)=ddx(A)

yp1(x)=0 (5)

Re-write equation (1) by neglecting the term sin2x ,

y+y2y=x (6)

Substitute equations (3), (4) and (5) in equation (6),

0+A2(Ax+B)=x (7)

Differentiate the equation (7) with respect to x,

ddx(A2(Ax+B))=ddx(x)ddx(A)+ddx(2Ax)+ddx(2B)=ddx(x)02A+0=12A=1

Simplify the equation,

A=12

Substitute 0 for x in equation (7),

0+A2(A(0)+B)=0

A2B=0 (8)

Substitute 12 for A in equation (8),

(12)2B=02B=12B=14

Substitute 12 for A and 14 for B in equation (3),

yp1(x)=12x14 (9)

Re-write equation (1) by neglecting x.

y+y2y=sin2x (10)

The particular solution yp2(x) is,

yp2(x)=Acos2x+Bsin2x (11)

Differentiate equation (11) with respect to x,

yp2(x)=ddx(Acos2x+Bsin2x)

yp2(x)=2Asin2x+2Bcos2x (12)

Differentiate equation (12) with respect to x,

yp2(x)=ddx(2Asin2x+2Bcos2x)

yp2(x)=4Acos2x4Bsin2x (13)

Substitute equations (11), (12), and (13) in equation (10),

(4Acos2x4Bsin2x)+(2Asin2x+2Bcos2x)2(Acos2x+Bsin2x)=sin2x(4A+2B2A)cos2x+(4B2A2B)sin2x=sin2x

(6A+2B)cos2x+(6B2A)sin2x=sin2x (14)

Substitute 0 for x in equation (14)

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Chapter 17 Solutions

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Sect-17.1 P-11ESect-17.1 P-12ESect-17.1 P-13ESect-17.1 P-14ESect-17.1 P-15ESect-17.1 P-16ESect-17.1 P-17ESect-17.1 P-18ESect-17.1 P-19ESect-17.1 P-20ESect-17.1 P-21ESect-17.1 P-22ESect-17.1 P-23ESect-17.1 P-24ESect-17.1 P-25ESect-17.1 P-26ESect-17.1 P-27ESect-17.1 P-28ESect-17.1 P-29ESect-17.1 P-30ESect-17.1 P-31ESect-17.1 P-32ESect-17.1 P-33ESect-17.1 P-34ESect-17.2 P-1ESect-17.2 P-2ESect-17.2 P-3ESect-17.2 P-4ESect-17.2 P-5ESect-17.2 P-6ESect-17.2 P-7ESect-17.2 P-8ESect-17.2 P-9ESect-17.2 P-10ESect-17.2 P-11ESect-17.2 P-12ESect-17.2 P-13ESect-17.2 P-14ESect-17.2 P-15ESect-17.2 P-16ESect-17.2 P-17ESect-17.2 P-18ESect-17.2 P-19ESect-17.2 P-20ESect-17.2 P-21ESect-17.2 P-22ESect-17.2 P-23ESect-17.2 P-24ESect-17.2 P-25ESect-17.2 P-26ESect-17.2 P-27ESect-17.2 P-28ESect-17.3 P-1ESect-17.3 P-2ESect-17.3 P-3ESect-17.3 P-4ESect-17.3 P-5ESect-17.3 P-6ESect-17.3 P-7ESect-17.3 P-8ESect-17.3 P-9ESect-17.3 P-10ESect-17.3 P-11ESect-17.3 P-12ESect-17.3 P-13ESect-17.3 P-14ESect-17.3 P-15ESect-17.3 P-16ESect-17.3 P-17ESect-17.3 P-18ESect-17.4 P-1ESect-17.4 P-2ESect-17.4 P-3ESect-17.4 P-4ESect-17.4 P-5ESect-17.4 P-6ESect-17.4 P-7ESect-17.4 P-8ESect-17.4 P-9ESect-17.4 P-10ESect-17.4 P-11ESect-17.4 P-12ECh-17 P-1RCCCh-17 P-2RCCCh-17 P-3RCCCh-17 P-4RCCCh-17 P-5RCCCh-17 P-1RQCh-17 P-2RQCh-17 P-3RQCh-17 P-4RQCh-17 P-1RECh-17 P-2RECh-17 P-3RECh-17 P-4RECh-17 P-5RECh-17 P-6RECh-17 P-7RECh-17 P-8RECh-17 P-9RECh-17 P-10RECh-17 P-11RECh-17 P-12RECh-17 P-13RECh-17 P-14RECh-17 P-15RECh-17 P-16RECh-17 P-17RECh-17 P-18RECh-17 P-19RECh-17 P-20RECh-17 P-21RE

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