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In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

et(y - t)dt +(1 +et)dy = 0

Short Answer

Expert verified

The solution isy=(t-1)et+C1+et.

Step by step solution

01

Evaluate whether the equation is exact

Hereety - tdt +1 +etdy = 0

The condition for exact is∂M∂y=∂N∂t.

M(t,y)=et(y-t)N(t,y)=(1+et)

∂M∂y=et=∂N∂t

This equation is exact.

02

Find the value of F (x, t)

Here

M(t,y)=et(y-t)F(t,y)=∫M(t,y)dt+g(y)=∫et(y-t)dt+g(y)=ety-ett+et+g(y)

03

Determine the value of g(y)

∂F∂y(t,y)=N(t,y)et+g'(y)=1+etg'(y)=1g(y)=y+C1

NowF(t,y) =ety -ett +et+ y +C1

The general solution of the differential equation isy=(t-1)et+C1+et

Hence the solution isrole="math" localid="1664172391787" y=(t-1)et+C1+et

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