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In Problems \(15 - 18\) verify that the indicated functionis an explicit solution of the given first-order differential equation. Proceed as in Example \(6\), by considering \(\phi \) simply as a function and give its domain. Then by considering \(\phi \) as a solution of the differential equation, give at least one interval \(I\) of definition.

\(y' = 25 + {y^2};y = 5tan5x\)

Short Answer

Expert verified

The indicated function is an explicit solution of the given differential equation and the interval is \(I:\left( { - \frac{\pi }{{10}},\frac{\pi }{{10}}} \right)\), and the domain is \(D:x \ne \frac{\pi }{{10}} + \frac{{n\pi }}{5}\).

Step by step solution

01

Define an explicit function.

Anexplicit solutionis one in which the dependent variable is expresseddirectlyin terms of the independent variable and constants.

Let the given function be .

Then, the first derivative of the function is,

\(y' = 25se{c^2}5x\)

02

Determine the explicit solution.

Substitute \(y\) and \(y'\) into the left-hand side of the differential equation.

\(\begin{aligned}{c}25se{c^2}5x &= 25 + 25ta{n^2}5x\\25se{c^2}5x &= 25\left( {1 + ta{n^2}5x} \right)\\25se{c^2}5x &= 25se{c^2}5x\end{aligned}\)

That is same as the right-hand side of the differential equation. The indicated function is an explicit solution of the given differential equation.

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Most popular questions from this chapter

In Problems \(31 - 34\) find values of m so that the function \(y = m{e^{mx}}\) is a solution of the given differential equation.

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