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Chapter 1: Q23E (page 1)

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

(ety+tety)dt+(tet+2)dy=0,y(0)=-1

Short Answer

Expert verified

The solution is .y=-2tet+2

Step by step solution

01

Evaluate the equation is exact

Here(ety+tety)dt+(tet+2)dy=0,y(0)=-1

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

M(t,y)=(ety+tety)N(t,y)=(tet+2)鈭�M鈭�y=et+tet=鈭�N鈭�t

This equation is exact.

02

Find the value of F(x, y)

Here

M(t,y)=(ety+tety)F(t,y)=鈭�M(t,y)dt+g(y)=鈭�(ety+tety)dt+g(y)=ytet+g(y)

03

Determine the value of g(y)

鈭�F鈭�y(t,y)=N(t,y)tet+g'(y)=(tet+2)g'(y)=2g(y)=2y+C1

NowF(t,y)=ytet+2y+C1

Therefore the solution of the differential equation is

ytet+2y=Cy=ctet+2

Apply the initial conditions.

y(0)=-1y(0)

-1=c0+2C=-2

The solutions isy=-2tet+2.

Hence the solution is y=-2tet+2

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Most popular questions from this chapter

Lunar Orbit. The motion of a moon moving in a planar orbit about a planet is governed by the equations d2xdt2=-Gmxr3,d2ydt2=-Gmyr3where r=(x2+y2)12, G is the gravitational constant, and m is the mass of the planet. Assume Gm = 1. When x(0)=1,x'(0)=y(0)=0,y'(0)=1the motion is a circular orbit of radius 1 and period 2蟺.

(a) The setting x1=x,x2=x',x3=y,x4=y'expresses the governing equations as a first-order system in normal form.

(b) Using localid="1664116258849" h=2蟺100鈮�0.0628318,compute one orbit of this moon (i.e., do N = 100 steps.). Do your approximations agree with the fact that the orbit is a circle of radius 1?

In Problems 9-13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship implicitly defines y as a function of x and use implicit differentiation.

x2-sin(x+y)=1,dydx=2xsec(x+y)-1

In problems 1-4Use Euler鈥檚 method to approximate the solution to the given initial value problem at the points x = 0.1, 0.2, 0.3, 0.4, and 0.5, using steps of size 0.1 (h = 0.1).

dydx=y(2-y),y(0)=3

In problems 1-4Use Euler鈥檚 method to approximate the solution to the given initial value problem at the points x = 0.1, 0.2, 0.3, 0.4, and 0.5, using steps of size 0.1 (h = 0.1).

dydx=x+y,y(0)=1

Question:Use a CAS to graphJ3/2(x),J-3/2(x),J5/2(x), and J-5/2(x).
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