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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.

\(3xy' + 5y = 0\)

Short Answer

Expert verified

There exists a constant solution (i.e.) \(y = 2\).

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,

\(\begin{array}{c}3x(0) + 5c = 10\\5c = 10\\c = 2\end{array}\)

Hence, the differential equation possesses a constant solution because the value of \(c\) satisfies the differential equation.

So, the value is \(y = 2\).

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In Problems 27–30 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.

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