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Chapter 8: Q56E (page 356)

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

\(5y' = 2y\)

Short Answer

Expert verified

The value of \(m\) is \(\frac{2}{5}\).

Step by step solution

01

Define a derivative of the function.

The derivatives of a function \(y = f{x)\) of a quantity x is a measurement of the rate at which the function's value y changes when the variable x varies. The derivatives of f with respect to x is what it's called.

Let the given function be\(5y' = 2y\). Then, the first derivative of the function is \(y' = m{e^{mx}}\).

02

Determine the value.

Substitute\(y\)and \(y'\)into the differential equation.

\(\begin{aligned} 5m{e^{mx}} &= 2{e^{mx}}\\{e^{mx}}(5m - 2) &= 0\\5m - 2 &= 0\\m &= \frac{2}{5}\end{aligned}\)

As \({e^{mx}}\) will never equal to \(0\), that implies that \(5m - 2 = 0\).

Thus, the value of \(m\) is \(\frac{2}{5}\).

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