Question

If \( \frac{dy}{dt} = 2y \), and the value of \( y \) at \( t = 0 \) is 2, then the value of \( y \) at \( t = 1 \) is ______ (rounded off to two decimal places).

Hint:

An equation of the form \( dy/dt = ky \) always has the solution \( y(t) = y_0 e^{kt} \), where \( y_0 \) is the initial value.

Answer 1

Correct Answer: 14.78

Step 1: Understanding the Concept:
We need to solve a first order ordinary differential equation from scratch using the initial condition given, rather than quoting a memorized formula.

Step 2: Key Formula or Approach:
The equation $\dfrac{dy}{dt} = 2y$ can be solved by the method of separation of variables, moving all the $y$ terms to one side and all the $t$ terms to the other.

Step 3: Detailed Explanation:
Separate the variables:
\[ \frac{dy}{y} = 2\,dt \]
Integrate both sides:
\[ \int \frac{1}{y}\,dy = \int 2\,dt \]
\[ \ln(y) = 2t + C \]
Apply the initial condition $y = 2$ at $t = 0$ to find $C$:
\[ \ln(2) = 2(0) + C \implies C = \ln(2) \]
So the particular solution is:
\[ \ln(y) = 2t + \ln(2) \]
Exponentiate both sides:
\[ y = e^{2t + \ln 2} = e^{2t} \cdot e^{\ln 2} = 2 e^{2t} \]
Now evaluate at $t = 1$:
\[ y(1) = 2 e^{2} = 2 \times 7.389056 = 14.778112 \]

Final Answer:
Rounded off to two decimal places, the value is 14.78.
\[ \boxed{y(1) = 14.78} \]