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Chapter 8: Problem 6

Does the function \(y(t)=6 e^{-3 t}\) satisfy the initial value problem \(y^{\prime}(t)-3 y(t)=0, y(0)=6 ?\)

Short Answer

Expert verified
Answer: Yes, the function \(y(t)=6e^{-3t}\) satisfies the initial value problem \(y'(t)-3y(t)=0, y(0)=6\).

Step by step solution

01

Compute the derivative of the function

To find out if the function satisfies the given IVP, we need to find its derivative. We compute the derivative of \(y(t)=6e^{-3t}\) with respect to \(t\) as follows: $$ y'(t) = \frac{d}{dt}(6e^{-3t}) $$ Using the chain rule, the derivative of the given function is $$ y'(t) = -18e^{-3t}. $$
02

Plug the function and its derivative into the differential equation

Now, we need to check if the function and its derivative satisfy the given differential equation \(y'(t)-3y(t)=0\). We plug in the values of y(t) and its derivative, y'(t), we obtained in step 1. $$ (-18e^{-3t}) - 3(6e^{-3t}) = 0 $$ Simplify the equation: $$ -18e^{-3t} - 18e^{-3t} = 0 $$ Combine the terms: $$ -36e^{-3t} = 0 $$
03

Determine if the equation holds true

The equation we obtained in step 2, \(-36e^{-3t} = 0\), does not hold true for all values of \(t\). However, we are interested in the initial value problem, which involves checking the given condition \(y(0) = 6\). So let's plug in the value, \(t = 0\), into the given function and see if it satisfies the initial condition.
04

Check the initial condition

We plug in \(t=0\) into the function \(y(t) = 6e^{-3t}\). $$ y(0) = 6e^{-3(0)} $$ Simplify the expression: $$ y(0) = 6e^0 $$ Since \(e^0 = 1\), we have: $$ y(0) = 6 $$
05

Conclude

We conclude that the given function \(y(t)=6e^{-3t}\) satisfies the given initial value problem \(y'(t)-3y(t)=0, y(0)=6\), because both the differential equation is satisfied in the general form and the initial condition is met.

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