Suppose y = 2e^-4x is the solution to the initial value problem y′ + ky = 0, y(0) = y₀. Find the value of y_₀.

Solution:

Given y = 2e^-4x is the solution to y′ + ky = 0 and also y(0) = y₀

y(0) = 2e^-4(0) = 2(1) = 2

But given y(0) = y₀⇒ y₀ = 2

Since y is the solution of the given equation, it must satisfy it

y′ + ky = 0

d(2e^-4x)/dx + k(2e^-4x) = 0

(-4)2e^-4x + k(2e^-4x)  =0

-8e^-4x +2ke^-4x = 0

Taking e^-4x as common and dividing,

⇒ -8 + 2k =0

⇒ 2k = 8

⇒ k = 4

Therefore, y₀= 2 and k = 4

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

Summary:

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