Question

in exercises 11 - 14, find the solution of the differential equation ( \frac{dy}{dt}=ky ), ( k ) a constant, that satisfies the given conditions.

1. ( k = 1.5 ), ( y(0)=100 ) 12. ( k=-0.5 ), ( y(0)=200 )
2. ( y(0)=50 ), ( y(5)=100 ) 14. ( y(0)=60 ), ( y(10)=30 )

Explanation:

Step1: Solve the differential equation

The differential equation \(\frac{dy}{dt}=ky\) is a first - order separable differential equation. We can rewrite it as \(\frac{dy}{y}=kdt\).
Integrating both sides:
\(\int\frac{dy}{y}=\int kdt\)
\(\ln|y| = kt + C\)
\(y = e^{kt + C}=e^{C}e^{kt}\). Let \(A = e^{C}\), then \(y = Ae^{kt}\).

Step2: Use the initial condition \(y(0)\)

Given \(y(0)=100\) and \(k = 1.5\). Substitute \(t = 0\) and \(y=100\) into \(y = Ae^{kt}\):
\(y(0)=Ae^{k\times0}\), so \(A=y(0)\). Since \(y(0) = 100\) and \(k = 1.5\), the solution is \(y = 100e^{1.5t}\).

Answer:

\(y = 100e^{1.5t}\)