Question

find the particular antiderivative of the following derivative that satisfies the given condition.
\\( \frac { d x } { d t } = 8 e ^ { t } - 5 ; \quad x ( 0 ) = 2 \\)
\\( x ( t ) = \square \\)

Explanation:

Step1: Integrate the derivative

We know that if \(\frac{dx}{dt}=8e^{t}-5\), then by the fundamental theorem of calculus \(x(t)=\int(8e^{t}-5)dt\).
Using the integral rules \(\int e^{t}dt = e^{t}+C_1\) and \(\int kdt=kt + C_2\) (\(k\) is a constant), we have:
\(x(t)=8\int e^{t}dt-5\int dt\)
\(x(t)=8e^{t}-5t + C\)

Step2: Use the initial condition

Given \(x(0) = 2\). Substitute \(t = 0\) and \(x=2\) into \(x(t)=8e^{t}-5t + C\):
\(2=8e^{0}-5\times0 + C\)
Since \(e^{0}=1\), we get \(2=8\times1-0 + C\)
\(C=2 - 8=-6\)

Answer:

\(x(t)=8e^{t}-5t - 6\)