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Chapter 12: Q39E (page 518)

In Problems \(37 - 40\) use the concept that \(y = c, - \infty < x < \infty \), is a constant function if and only if \(y' = 0\) to determine whether the given differential equation possesses constant solutions.

\((y - 1)y' = 1\)

Short Answer

Expert verified

There is no constant solution.

Step by step solution

01

Define constant solution of the function.

When the derivative of a differential equation is zero, the constant solutions appear. If \(g(a) = 0\) for some \(a\), then \(y(t) = a\) is a constant solution of the equation, because \(y = f(t)g(a) = 0\).

02

Determine whether it has constant solution.

Since\(y = c\), implies that \(y' = 0\), then the differential equation becomes,

\(\begin{aligned}{c}(c - 1)(0) = 1\\0 \ne 1\end{aligned}\)

Hence, the differential equation does not possess a constant solution because the value of \(c\) becomes zero.

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Most popular questions from this chapter

In Problems 27鈥�30 use (12) of Section 1.1 to verify that the indicated function is a solution of the given differential equation. Assume an appropriate interval I of definition of each solution.

\(\frac{{dy}}{{dx}} - 2xy = {e^{{x_z}}}; y = {e^{{x^2}}}\int_0^x {{e^{t - {t^2}}}} dt\)

In Problems \(37 - 40\) use the concept that \(y = c, - \infty < x < \infty \), is a constant function if and only if \(y' = 0\) to determine whether the given differential equation possesses constant solutions.

\(y'' + 4y' + 6y = 10\)

In Problems \(23 - 26\) verify that the indicated function is an explicit solution of the given differential equation. Give an interval of definition \(I\) for each solution.

\({x^2}y'' + xy' + y = sec(lnx); y = cos(lnx)ln(cos(lnx)) + (lnx)sin(lnx)\)

(a) Verify that the one-parameter family \({y^2} - 2y = {x^2} - x + c\) is an implicit solution of the differential equation \((2y - 2)y' = 2x - 1\).

(b) Find a member of the one-parameter family in part (a) that satisfies the initial condition \(y(0) = 1\).

(c) Use your result in part (b) to and an explicit function \(y = \phi (x)\) that satisfies \(y(0) = 1\). Give the domain of the function \(\phi \). Is \(y = \phi (x)\) a solution of the initial-value problem? If so, give its interval \(I\) of definition; if not, explain.

In Problems \(1 - 8\) state the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear by matching it with \((6)\).

\(\ddot x - \left( {1 - \frac{{{{\dot x}^2}}}{3}} \right)\dot x + x = 0\)
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