Solve the initial value problem: $$ \frac{d y}{d x}+3 y=8, \quad y(0)=0 $$ $$ y(x)= $$

Question

UpStudy AI · Accepted Answer

**Step 1. Write the differential equation in standard form**

The given differential equation is

$$
\frac{dy}{dx} + 3y = 8
$$

with the initial condition

$$
y(0) = 0.
$$

---

**Step 2. Find the integrating factor**

The integrating factor $$\mu(x)$$ is given by:

$$
\mu(x) = e^{\int 3\,dx} = e^{3x}.
$$

---

**Step 3. Multiply the equation by the integrating factor**

Multiply both sides of the differential equation by $$e^{3x}$$:

$$
e^{3x}\frac{dy}{dx} + 3e^{3x}y = 8e^{3x}.
$$

The left-hand side becomes the derivative of $$e^{3x}y$$:

$$
\frac{d}{dx}\left(e^{3x}y\right) = 8e^{3x}.
$$

---

**Step 4. Integrate both sides**

Integrate the equation with respect to $$x$$:

$$
\int \frac{d}{dx}\left(e^{3x}y\right) dx = \int 8e^{3x} dx.
$$

This gives:

$$
e^{3x}y = \frac{8}{3}e^{3x} + C,
$$

where $$C$$ is the constant of integration.

---

**Step 5. Solve for $$y$$ and apply the initial condition**

Divide both sides by $$e^{3x}$$:

$$
y = \frac{8}{3} + Ce^{-3x}.
$$

Now, use the initial condition $$y(0) = 0$$:

$$
0 = \frac{8}{3} + C \cdot e^{0} \quad \Rightarrow \quad 0 = \frac{8}{3} + C.
$$

Thus,

$$
C = -\frac{8}{3}.
$$

Substitute the value of $$C$$ back into the equation for $$y$$:

$$
y = \frac{8}{3} - \frac{8}{3}e^{-3x}.
$$

---

**Step 6. Write the final answer**

$$
\boxed{y(x) = \frac{8}{3}\left(1 - e^{-3x}\right)}
$$
$$ y(x) = \frac{8}{3}\left(1 - e^{-3x}\right) $$

Solve the initial value problem: \( \frac{d y}{d x}+3 y=8, \quad y(0)=0 \) \( y(x)= \)