8. Solve the following initial value problem: $$ \frac{d y}{d t}=y(y-4), \quad y(0)=-2 $$ (a) $$ y(t)=2-2 \sqrt{1+3 e^{4 t}} \quad y(t)=\frac{4}{1+C e^{4 t}} $$ (b) $$ y(t)=\frac{4}{1-3 e^{-4 t}} $$ (c) $$ y(t)=2-2 \sqrt{1+3 e^{-4 t}} $$ (d) $$ y(t)=\frac{4}{1-3 e^{4 t}} $$

Question

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To solve the initial value problem:

$$
\frac{dy}{dt} = y(y - 4), \quad y(0) = -2
$$

**Step 1: Separate Variables**

$$
\frac{dy}{y(y - 4)} = dt
$$

**Step 2: Partial Fraction Decomposition**

We can decompose the left-hand side using partial fractions:

$$
\frac{1}{y(y - 4)} = \frac{A}{y} + \frac{B}{y - 4}
$$

Solving for $$ A $$ and $$ B $$:

$$
1 = A(y - 4) + By
$$

Setting $$ y = 0 $$:

$$
1 = A(-4) \implies A = -\frac{1}{4}
$$

Setting $$ y = 4 $$:

$$
1 = B(4) \implies B = \frac{1}{4}
$$

So,

$$
\frac{1}{y(y - 4)} = -\frac{1}{4y} + \frac{1}{4(y - 4)}
$$

**Step 3: Integrate Both Sides**

$$
\int \left( -\frac{1}{4y} + \frac{1}{4(y - 4)} \right) dy = \int dt
$$

$$
-\frac{1}{4} \ln |y| + \frac{1}{4} \ln |y - 4| = t + C
$$

Multiply through by 4:

$$
-\ln |y| + \ln |y - 4| = 4t + C'
$$

Combine logarithms:

$$
\ln \left| \frac{y - 4}{y} \right| = 4t + C'
$$

Exponentiate both sides:

$$
\frac{y - 4}{y} = Ce^{4t}
$$

Solving for $$ y $$:

$$
y - 4 = Ce^{4t}y
$$

$$
y(1 - Ce^{4t}) = 4
$$

$$
y = \frac{4}{1 - Ce^{4t}}
$$

**Step 4: Apply Initial Condition**

Using $$ y(0) = -2 $$:

$$
-2 = \frac{4}{1 - C}
$$

$$
-2(1 - C) = 4
$$

$$
-2 + 2C = 4
$$

$$
2C = 6 \implies C = 3
$$

**Final Solution:**

$$
y(t) = \frac{4}{1 - 3e^{4t}}
$$

**Matching with the Options:**

The correct solution corresponds to option **(d)**.

**Answer:**  
Problem 8(d) Answer

$$ y(t)=\frac{4}{1-3 e^{4 t}} $$
$$ y(t) = \frac{4}{1 - 3e^{4t}} $$

Solve the following initial value problem:

(a)
(b)
©
(d)

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Solve the following initial value problem: (a) (b) © (d)