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Chapter 2: Q26E (page 64)

In Problems 21鈥�26, solve the initial value problem.

(tany-2)dx+(xsec2y+1y)dy=0,y(0)=1

Short Answer

Expert verified

The solution is (tany-2)x+lny=0.

Step by step solution

01

Evaluate the equation is exact

Here (tany-2)dx+xsec2y+1ydy=0,y(0)=1

The condition for exact is 鈭�M鈭�y=鈭�N鈭�x.

M(x,y)=(tany-2)N(x,y)=xsec2y+1y鈭�M鈭�y=sec2y=鈭�N鈭�x

This equation is exact.

02

Find the value of F(x, y)

Here

M(x,y)=(tany-2)F(x,y)=鈭�M(x,y)dx+g(y)=鈭�(tany-2)dx+g(y)=(tany-2)x+g(y)

03

Determine the value of g(y)

鈭�F鈭�y(x,y)=N(x,y)xsec2y+g'(y)=xsec2y+1yg'(y)=1yg(y)=lny

Now F(x,y)=(tany-2)x+lny

Therefore, the solution of the differential equation is

(tany-2)x+lny=C

Apply the initial conditionsy(0)=1.

(tan1-2)0+ln1=CC=0

The solutions is (tany-2)x+lny=0.

Hence the solution is (tany-2)x+lny=0

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