Solve the initial-value problem 
y
′
=
ln
⁡
(
2
t
+
2
)
y 
′
 =ln(2t+2) with 
y
(
0
)
=
2
y(0)=2. 

y
(
t
)
=
y(t)=

We are tasked with solving the initial value problem:

\[
y' = \ln(2t + 2), \quad y(0) = 2
\]

### Step 1: Integrate both sides
We start by integrating the right-hand side with respect to \( t \) to find \( y(t) \). 

\[
y(t) = \int \ln(2t + 2) \, dt
\]

To integrate \( \ln(2t + 2) \), we can use integration by parts. Let:

\[
u = \ln(2t + 2), \quad dv = dt
\]

Then, the derivative of \( u \) and the integral of \( dv \) are:

\[
du = \frac{2}{2t + 2} \, dt, \quad v = t
\]

Now applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:

\[
\int \ln(2t + 2) \, dt = t \ln(2t + 2) - \int t \cdot \frac{2}{2t + 2} \, dt
\]

Simplifying the second integral:

\[
\int \frac{2t}{2t + 2} \, dt = \int \left(1 - \frac{2}{2t + 2}\right) \, dt = t - \ln(2t + 2)
\]

Thus, we have:

\[
\int \ln(2t + 2) \, dt = t \ln(2t + 2) - (t - \ln(2t + 2)) = t \ln(2t + 2) - t + \ln(2t + 2) + C
\]

### Step 2: Apply the initial condition
Now, we apply the initial condition \( y(0) = 2 \). Substituting \( t = 0 \) into the general solution:

\[
y(0) = 0 \ln(2(0) + 2) - 0 + \ln(2(0) + 2) + C = \ln(2) + C
\]

We know that \( y(0) = 2 \), so:

\[
2 = \ln(2) + C
\]

Solving for \( C \):

\[
C = 2 - \ln(2)
\]

### Step 3: Final solution
Thus, the solution to the initial value problem is:

\[
y(t) = t \ln(2t + 2) - t + \ln(2t + 2) + 2 - \ln(2)
\]

This is the general solution to the problem.

Would you like more details, or do you have any questions?

Here are 5 related questions:
1. How does integration by parts work in general?
2. What are other methods to solve differential equations?
3. How would the solution change if the initial condition was different?
4. Can this problem be solved using numerical methods?
5. How can we approximate the behavior of \( y(t) \) for large \( t \)?

**Tip:** When solving differential equations, always check if an initial condition is provided; it helps determine the constant of integration.

Solve the initial-value problem y' = ln(2t + 2) with y(0) = 2.

Learn how to solve the initial-value problem y' = ln(2t + 2) with y(0) = 2 using integration by parts. This step-by-step solution covers key concepts in differential equations and integration, suitable for undergraduate students.

Math Problem Statement

Solve the initial-value problem y ′

ln ⁡ ( 2 t + 2 ) y ′ =ln(2t+2) with y ( 0 )

y ( t )

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation