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Chapter 1: Q24E (page 1)

In Problems 21–26, solve the initial value problem (etx+1)dt+(et-1)dx=0,x(1)=1

Short Answer

Expert verified

The solution isx=e-tet-1

Step by step solution

01

Evaluate the equation is exact

Here(etx+1)dt+(et-1)dx=0,x(1)=1

The condition for exact is∂M∂x=∂N∂t .

M(x,t)=(etx+1)N(x,t)=(et-1)

∂M∂x=et=∂N∂t

This equation is exact.

02

Find the value of F(x,t)

Here

M(t,y)=(etx+1)F(x,t)=∫M(x,t)dt+g(x)=∫(etx+1)dt+g(x)=(etx+t)+g(x)
03

Determine the value of g(y)

∂F∂x(x,t)=N(x,t)et+g'(x)=(et-1)g'(x)=-1g(x)=-x+C1

NowF(x,t)=etx+t-x+C1

Therefore, the solution of the differential equation is

etx+t-x=Cx=c-tet-1

Apply the initial conditionsx(1)=1x(1)=1 .

1=c-1e1-1C=e

The solutions is x=e-tet-1.

Hence the solution is x=e-tet-1

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