Question

Suppose y=2e−4x is the solution to the initial value problem y′+ky=0,y(0)=y0. Find the value of y0

Answer 1

Answer on Question #69660 – Math – Differential Equations

Question

Suppose $y = 2e^{-4x}$ is the solution to the initial value problem

$$y' + ky = 0, \quad y(0) = y_0$$

Find the value of $y_0$.

Solution

We have that

$$y(0) = 2e^{-4 \cdot 0} = 2$$

Then

$$y_0 = 2.$$

The given function satisfies the equation

$$y' + ky = 0$$

That is

$$\begin{array}{l} (2e^{-4x})' + k(2e^{-4x}) = 0 \\ -8e^{-4x} + k(2e^{-4x}) = 0 \\ 2k = 8 \\ k = 4 \\ \end{array}$$

Answer: $y_0 = 2, k = 4$.

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