To solve the initial value problem $ y' = 3e^{3x - y} $ with $ y(0) = \ln(4) $, follow these steps:

1. **Rewrite the differential equation:**
$$
\frac{dy}{dx} = 3e^{3x - y}
$$

2. **Separate the variables:**
$$
e^y \, dy = 3e^{3x} \, dx
$$

3. **Integrate both sides:**
$$
\int e^y \, dy = \int 3e^{3x} \, dx
$$

4. **Solve the integrals:**
$$
\int e^y \, dy = e^y + C_1
$$
$$
\int 3e^{3x} \, dx = e^{3x} + C_2
$$

5. **Combine the constants of integration:**
$$
e^y = e^{3x} + C
$$
where $ C = C_2 - C_1 $.

6. **Apply the initial condition $ y(0) = \ln(4) $:**
$$
e^{\ln(4)} = e^{3 \cdot 0} + C
$$
$$
4 = 1 + C
$$
$$
C = 3
$$

7. **Substitute $ C $ back into the equation:**
$$
e^y = e^{3x} + 3
$$

8. **Solve for $ y $:**
$$
y = \ln(e^{3x} + 3)
$$

Thus, the solution to the initial value problem is:
$$
y = \ln(e^{3x} + 3)
$$

Question

To solve the initial value problem $ y' = 3e^{3x - y} $ with $ y(0) = \ln(4) $, follow these steps:

1. **Rewrite the differential equation:**
   $$
   \frac{dy}{dx} = 3e^{3x - y}
   $$

2. **Separate the variables:**
   $$
   e^y \, dy = 3e^{3x} \, dx
   $$

3. **Integrate both sides:**
   $$
   \int e^y \, dy = \int 3e^{3x} \, dx
   $$

4. **Solve the integrals:**
   $$
   \int e^y \, dy = e^y + C_1
   $$
   $$
   \int 3e^{3x} \, dx = e^{3x} + C_2
   $$

5. **Combine the constants of integration:**
   $$
   e^y = e^{3x} + C
   $$
   where $ C = C_2 - C_1 $.

6. **Apply the initial condition $ y(0) = \ln(4) $:**
   $$
   e^{\ln(4)} = e^{3 \cdot 0} + C
   $$
   $$
   4 = 1 + C
   $$
   $$
   C = 3
   $$

7. **Substitute $ C $ back into the equation:**
   $$
   e^y = e^{3x} + 3
   $$

8. **Solve for $ y $:**
   $$
   y = \ln(e^{3x} + 3)
   $$

Thus, the solution to the initial value problem is:
$$
y = \ln(e^{3x} + 3)
$$

Knowee AI · Accepted Answer