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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.

7. y" – 2y' + 5y = sin x, y(0) = 1, y'(0) = 1

To determine

To solve: The differential equation by the method of undetermined coefficients.

Explanation

Given data:

The differential equation is,

y+2y+5y=sinx (1)

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

Consider the auxiliary equation is,

r22r+5=0 (2)

Roots of equation (2) are,

r=2±(2)24(1)(5)2(1){r=b±b24ac2afortheequationofar2+br+c=0}=1±2i

Write the expression for the complementary solution,

yc(x)=ex(c1cos2x+c2sin2x)

The particular solution yp(x) is,

yp(x)=Acosx+Bsinx (3)

Differentiate equation (3) with respect to x,

yp(x)=ddx(Acosx+Bsinx)

yp(x)=Asinx+Bcosx (4)

Differentiate equation (3) with respect to x,

yp(x)=ddx(Asinx+Bcosx)

yp(x)=AcosxBsinx (5)

Substitute equations (3), (4) and (5) in equation (1),

AcosxBsinx+2(Asinx+Bcosx)+5(Acosx+Bsinx)=sinx

(4A2B)cosx+(2A+4B)sinx=sinx (6)

Substitute 0 for x in equation (6),

(4A2B)cos0+(2A+4B)sin0=sin0(4A2B)(1)+0=0

4A2B=0 (7)

Substitute 90° for x in equation (6),

(4A2B)cos90°+(2A+4B)sin90°=sin90°(4A2B)(0)+(2A+4B)(1)=(1)

2A+4B=1 (8)

Multiply 2 on both sides of equation (8),

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