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Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378
Chapter 16, Problem 50RE
Textbook Problem
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Using Initial Conditions In Exercises 45-50, solve the differential equation by the method of undetermined coefficients subject to the initial conditions.

y ' ' ' y ' ' = 4 x 2

y ( 0 ) = 1 , y ' ( 0 ) = 1 , y ' ' ( 0 ) = 1

To determine

To calculate: The solution of the given differential equation y'''y''=4x2 by the method of undetermined coefficients subject to the initial conditions y(0)=1,y'(0)=1,y''(0)=1.

Explanation of Solution

Given information:

The given differential equation is y'''y''=4x2 and the initial conditions are y(0)=1,y'(0)=1,y''(0)=1.

Concept Used:

If yp is the particular solution of the differential equation y''+ay'+by=F(x) and yh is the general solution, then y=yh+yp is the general solution of the non homogeneous equation.

Calculation:

The characteristics equation is m3m2=0.

First, we have to find the solution of the characteristics equation m3m2=0.

m3m2=0m2(m1)=0m=0orm=1

Therefore,

yh=C1+C2x+C3x2ex

Let yp be generalized form of 4x2 .

yp=Ax2+Bx3+Cx4yp'=2Ax+3Bx2+4Cx3yp''=2A+6Bx+12Cx2yp'''=6B+24Cx

Substituting in the original equation yields

y'''y''=4x26B+24Cx2A6Bx12Cx2=4x2

Equating the coefficient of like terms yields

12C=4C=1324C6B=024C=6BB=4CB=432A+6B=02A=6BA=3BA=4

So,

A=4,B=43 and C=13

Therefore, yp=4x243x313x4 and the general solution is ,

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Chapter 16 Solutions

Multivariable Calculus
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