Solve the initial value problem $$ \frac{d u}{d t}=e^{5 u+8 t}, \quad u(0)=12 $$ $$ u(t)= $$

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We start with the differential equation
$$
\frac{d u}{d t}=e^{5u+8t},
$$
with the initial condition
$$
u(0)=12.
$$

**Step 1. Separate Variables**

We can rewrite the right‐hand side as
$$
e^{5u+8t}=e^{5u}e^{8t}.
$$
Thus, the equation becomes
$$
\frac{du}{dt}=e^{5u}e^{8t}.
$$
Separate the variables by dividing by $$ e^{5u} $$ and multiplying by $$ dt $$:
$$
e^{-5u}\,du = e^{8t}\,dt.
$$

**Step 2. Integrate Both Sides**

Integrate both sides:
$$
\int e^{-5u}\,du = \int e^{8t}\,dt.
$$
The integrals are computed as follows:
$$
\int e^{-5u}\,du = -\frac{1}{5}e^{-5u} + C_1
$$
and
$$
\int e^{8t}\,dt = \frac{1}{8}e^{8t} + C_2.
$$
We can combine the constants of integration into a single constant $$ C $$. Thus, we write:
$$
-\frac{1}{5}e^{-5u} = \frac{1}{8}e^{8t} + C.
$$

**Step 3. Solve for the Constant $$ C $$**

Multiply both sides by $$-5$$ to simplify:
$$
e^{-5u} = -\frac{5}{8}e^{8t} -5C.
$$
We rename $$-5C$$ as a new constant $$ K $$:
$$
e^{-5u} = K - \frac{5}{8}e^{8t}.
$$
Now, apply the initial condition $$ u(0)=12 $$:
$$
e^{-5\cdot12} = e^{-60} = K - \frac{5}{8}e^{0} = K - \frac{5}{8}.
$$
Thus,
$$
K = e^{-60} + \frac{5}{8}.
$$

**Step 4. Write the General Solution**

Substitute the constant $$ K $$ back into the equation:
$$
e^{-5u} = e^{-60} + \frac{5}{8} - \frac{5}{8}e^{8t}.
$$
Factor the constant term:
$$
e^{-5u} = e^{-60} + \frac{5}{8}(1-e^{8t}).
$$

**Step 5. Solve for $$ u(t) $$**

Take the natural logarithm of both sides:
$$
-5u = \ln\Bigl(e^{-60} + \frac{5}{8}(1-e^{8t})\Bigr).
$$
Thus, the solution is
$$
u(t) = -\frac{1}{5}\ln\left(e^{-60}+\frac{5}{8}\bigl(1-e^{8t}\bigr)\right).
$$

The final solution to the initial value problem is
$$
\boxed{u(t) = -\frac{1}{5}\ln\left(e^{-60}+\frac{5}{8}\bigl(1-e^{8t}\bigr)\right)}.
$$
The solution to the initial value problem is: $$ u(t) = -\frac{1}{5}\ln\left(e^{-60}+\frac{5}{8}\bigl(1-e^{8t}\bigr)\right) $$

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