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Chapter 1: Q14E (page 1)

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

If it is, then solve it.

(ty)dy+(1+lny)dt=0

Short Answer

Expert verified

The solution is tlny+t=C.

Step by step solution

01

Evaluate whether the equation is exact

Heretydy+(1+lny)dt=0

The condition for exact is∂M∂t=∂N∂y.

M(y,t)=tyN(y,t)=1+lny

∂M∂y=1y=∂N∂t

This equation is exact.

02

Find the value of F(y, t)

Here

M(y,t)=tyF(y,t)=∫M(y,t)dy+g(t)=∫tydy+g(t)=tlny+g(t)

03

determine the value of g (t)

NowFy,t= tlny + t +C1

The general solution of the differential equation is tlny+t=C.

Hence the solution istlny+t=C

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Most popular questions from this chapter

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem

dudx=3u-4v;u(0)=1'dvdx=2u-3v;v(0)=1

at x = 1. Starting with h=1, continue halving the step size until two successive approximations of u(1)and v(1) differ by at most 0.001.

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

If it is, then solve it.

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

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.

exy+y=x-1,dydx=e-xy-ye-xy+x

(a) For the initial value problem (12) of Example 9. Show that ϕ1(x)=0 andϕ2(x)=(x-2)3are solutions. Hence, this initial value problem has multiple solutions. (See also Project G in Chapter 2.)

(b) Does the initial value problemy'=3y23,y(0)=10-7have a unique solution in a neighbourhood ofx=0?

Verify thatx2+cy2=1, where c is an arbitrary non-zero constant, is a one-parameter family of implicit solutions todydx=xyx2-1 and graph several of the solution curves using the same coordinate axes.
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