Solving the Differential Equation dy/dt = 2y

We are given a first-order linear differential equation and an initial condition:

• Differential Equation: $\frac{dy}{dt} = 2y$
• Initial Condition: $y(0) = 2$

We need to find the value of $y$ when $t = 1$.

Separating Variables

To solve the differential equation, we first separate the variables ($y$ and $t$):

$ \frac{dy}{y} = 2 dt $

Integrating Both Sides

Next, we integrate both sides of the equation:

$ \int \frac{dy}{y} = \int 2 dt $

This yields:

$ \ln|y| = 2t + C_1 $

where $C_1$ is the constant of integration.

Finding the General Solution

To find $y$, we exponentiate both sides:

$ |y| = e^{2t + C_1} $

$ |y| = e^{C_1} e^{2t} $

Let $C = \pm e^{C_1}$. This gives the general solution:

$ y(t) = C e^{2t} $

Applying the Initial Condition

We use the initial condition $y(0) = 2$ to find the value of the constant $C$:

$ 2 = C e^{2(0)} $

$ 2 = C e^0 $

$ 2 = C $

Determining the Particular Solution

Substituting $C=2$ back into the general solution, we get the particular solution:

$ y(t) = 2 e^{2t} $

Calculating y at t = 1

Now, we evaluate $y(t)$ at $t = 1$:

$ y(1) = 2 e^{2(1)} $

$ y(1) = 2 e^2 $

Using the approximate value $e \approx 2.71828$:

$ e^2 \approx (2.71828)^2 \approx 7.389056 $

$ y(1) \approx 2 \times 7.389056 $

$ y(1) \approx 14.778112 $

Rounding the Result

Rounding the value of $y(1)$ to two decimal places:

$ y(1) \approx 14.78 $

This value falls within the range specified in the answer.