Question

In Exercises \(11-14\) , find the solution of the differential equation \(d y / d t=k y, k\) a constant, that satisfies the given conditions. $$k=1.5, \quad y(0)=100$$

Short answer

The solution to the differential equation \(\frac{dy}{dt} = 1.5y\) that satisfies the condition \(y(0)=100\) is \(y = 100e^{1.5t}\).

Step-by-step solution

Separate variables and integrate

Rewrite the differential equation \( \frac{dy}{dt} = 1.5y \) by separating 'y' and 't' on different sides: \( \frac{1}{y} dy = 1.5 dt \). Integrate both sides of the equation with respect to their variables: \( \int \frac{1}{y} dy = \int 1.5 dt \) which gives \( ln|y| = 1.5t + C \), where 'C' is the constant of integration.

Express 'y' in exponentiated form

Apply the exponential function to both sides to get rid of the natural logarithm: \(e^{ln|y|} = e^{1.5t+C}\). This simplifies to \(|y| = e^{1.5t+C}\), or even further as \(y = ±e^{C} \cdot e^{1.5t}\). Since 'y' is the size of a population which is always positive, we ignore the negative solution.

Apply the initial condition

Substitute the given initial condition \( y(0)=100 \) into the equation to find 'C'. This leads to \(100 = ±e^{C} \cdot e^{0}\), which simplifies to \(100 = ±e^{C}\). Solving for 'C', we get 'C'= ln(100), or about 4.605. Thus, \(y = e^{4.605} \cdot e^{1.5t}\).

Simplify the function

Finally, simplify the equation. Given that \(e^{4.605}\) is approximately equal to 100, the solution to the differential equation that satisfies the given condition is \(y = 100e^{1.5t}\).