Question: Find the general solution of the differential equation. Then, use the initial condition to find the corresponding particular solution.
y' - 3y = 7e^{3x}, y(0) = 0
The general solution is y = [ ].
The particular solution for y(0) = 0 is y = [ ].
1. **State the problem:**
We need to solve the first-order linear differential equation:
$$y' - 3y = 7e^{3x}$$
with the initial condition:
$$y(0) = 0$$
2. **Identify the type of equation and formula:**
This is a linear differential equation of the form:
$$y' + p(x)y = q(x)$$
where here, $$p(x) = -3$$ and $$q(x) = 7e^{3x}$$.
The integrating factor (IF) is given by:
$$\mu(x) = e^{\int p(x) dx} = e^{\int -3 dx} = e^{-3x}$$
3. **Multiply both sides by the integrating factor:**
$$e^{-3x} y' - 3 e^{-3x} y = 7 e^{3x} e^{-3x}$$
which simplifies to:
$$e^{-3x} y' - 3 e^{-3x} y = 7$$
4. **Recognize the left side as a derivative:**
The left side is the derivative of $$y e^{-3x}$$:
$$\frac{d}{dx} \left(y e^{-3x}\right) = 7$$
5. **Integrate both sides with respect to $$x$$:**
$$\int \frac{d}{dx} \left(y e^{-3x}\right) dx = \int 7 dx$$
which gives:
$$y e^{-3x} = 7x + C$$
where $$C$$ is the constant of integration.
6. **Solve for $$y$$:**
$$y = e^{3x} (7x + C)$$
This is the **general solution**.
7. **Apply the initial condition $$y(0) = 0$$:**
Substitute $$x=0$$ and $$y=0$$:
$$0 = e^{0} (7 \cdot 0 + C) = 1 \cdot C = C$$
So, $$C = 0$$.
8. **Write the particular solution:**
$$y = 7x e^{3x}$$
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**Final answers:**
- General solution: $$y = e^{3x} (7x + C)$$
- Particular solution for $$y(0) = 0$$: $$y = 7x e^{3x}$$