# 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) = y_{0}

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

But given y(0) = y_{0}⇒ y_{0} = 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, y_{0}= 2 and k = 4

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

**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.

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