3. (2 points) Solve the following initial value problem: $$ \frac{d y}{d t}=2 t e^{2 y}, y(0)=1 $$ (a) $$ y(t)=\frac{1}{2} \ln \left(e^{2}+2 t^{2}\right) $$ (b) $$ y(t)=-\frac{1}{2} \ln \left(e^{-2}+\frac{1}{2} t^{2}\right) $$ (c) $$ y(t)=-\frac{1}{2} \ln \left(e^{-2}-2 t^{2}\right) $$ (d) $$ y(t)=-\frac{1}{2} \ln \left(e^{-2}-\frac{1}{2} t^{2}\right) $$

Question

UpStudy AI · Accepted Answer

To solve the initial value problem:

$$
\frac{dy}{dt} = 2t e^{2y}, \quad y(0) = 1
$$

**Step 1: Separate Variables**

Rearrange the equation to separate $$ y $$ and $$ t $$:

$$
\frac{dy}{e^{2y}} = 2t \, dt
$$

**Step 2: Integrate Both Sides**

Integrate both sides:

$$
\int e^{-2y} \, dy = \int 2t \, dt
$$

The integrals are:

$$
-\frac{1}{2} e^{-2y} = t^2 + C
$$

**Step 3: Solve for $$ y $$**

Multiply both sides by $$-2$$:

$$
e^{-2y} = -2t^2 + C'
$$

Apply the initial condition $$ y(0) = 1 $$:

$$
e^{-2(1)} = C' \Rightarrow e^{-2} = C'
$$

So, the equation becomes:

$$
e^{-2y} = e^{-2} - 2t^2
$$

Take the natural logarithm on both sides:

$$
-2y = \ln(e^{-2} - 2t^2)
$$

Finally, solve for $$ y $$:

$$
y = -\frac{1}{2} \ln(e^{-2} - 2t^2)
$$

This corresponds to option **(c)**.

**Answer:**  
Answer (c) $$ y(t)=-\frac{1}{2} \ln \left(e^{-2}-2 t^{2}\right) $$
The solution is $$ y(t) = -\frac{1}{2} \ln \left(e^{-2} - 2t^{2}\right) $$, which is option (c).

(2 points) Solve the following initial value problem:

(a)
(b)
©
(d)

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