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).
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).
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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.
Updated On: Aug 5, 2026
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Correct Answer: 14.78
Solution and Explanation
Step 1: Understanding the Concept:
This is a first-order linear differential equation of the growth/decay type.
Any equation written as \( \frac{dy}{dt} = ky \) describes exponential growth (if k is positive) and always has the same standard solution form.
Step 2: Key Formula or Approach:
For \( \frac{dy}{dt} = ky \) with initial condition \( y(0) = y_0 \), the solution is always:
\[ y(t) = y_0 \, e^{kt} \]
Here we can read off \( k = 2 \) directly from the given equation, and \( y_0 = 2 \) from the initial condition, without needing to integrate from scratch.
Step 3: Detailed Explanation:
Substitute \( y_0 = 2 \) and \( k = 2 \) into the standard solution:
\[ y(t) = 2 e^{2t} \]
Now evaluate at \( t = 1 \):
\[ y(1) = 2 e^{2} \]
Using \( e \approx 2.718281828 \), we get \( e^2 \approx 7.389056 \).
\[ y(1) = 2 \times 7.389056 = 14.778112 \]
Final Answer:
Rounded off to two decimal places, \( y(1) = 14.78 \).
\[ \boxed{y(1) = 14.78} \]
This is a first-order linear differential equation of the growth/decay type.
Any equation written as \( \frac{dy}{dt} = ky \) describes exponential growth (if k is positive) and always has the same standard solution form.
Step 2: Key Formula or Approach:
For \( \frac{dy}{dt} = ky \) with initial condition \( y(0) = y_0 \), the solution is always:
\[ y(t) = y_0 \, e^{kt} \]
Here we can read off \( k = 2 \) directly from the given equation, and \( y_0 = 2 \) from the initial condition, without needing to integrate from scratch.
Step 3: Detailed Explanation:
Substitute \( y_0 = 2 \) and \( k = 2 \) into the standard solution:
\[ y(t) = 2 e^{2t} \]
Now evaluate at \( t = 1 \):
\[ y(1) = 2 e^{2} \]
Using \( e \approx 2.718281828 \), we get \( e^2 \approx 7.389056 \).
\[ y(1) = 2 \times 7.389056 = 14.778112 \]
Final Answer:
Rounded off to two decimal places, \( y(1) = 14.78 \).
\[ \boxed{y(1) = 14.78} \]
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