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The initial value problem provided is:

\[
t^2 \frac{dx}{dt} + 5tx = t^4 \ln t + 4, \quad x(1) = 0.
\]

This is a first-order linear differential equation in \( x(t) \). To solve it, we'll follow the steps for solving linear differential equations:

### Step 1: Write the equation in standard form
First, rewrite the equation:

\[
\frac{dx}{dt} + \frac{5x}{t} = t^2 \ln t + \frac{4}{t^2}.
\]

### Step 2: Find the integrating factor
The integrating factor \( \mu(t) \) is found using the formula:

\[
\mu(t) = e^{\int \frac{5}{t} dt} = e^{5 \ln t} = t^5.
\]

### Step 3: Multiply the equation by the integrating factor
Multiply both sides of the equation by \( t^5 \):

\[
t^5 \frac{dx}{dt} + 5t^4 x = t^7 \ln t + 4t^3.
\]

The left-hand side is now the derivative of \( t^5 x \):

\[
\frac{d}{dt} (t^5 x) = t^7 \ln t + 4t^3.
\]

### Step 4: Integrate both sides
Integrate both sides with respect to \( t \):

\[
t^5 x = \int (t^7 \ln t + 4t^3) dt.
\]

We now compute the integrals:

\[
\int t^7 \ln t \, dt \quad \text{and} \quad \int 4t^3 \, dt.
\]

1. For \( \int t^7 \ln t \, dt \), we use integration by parts, letting \( u = \ln t \) and \( dv = t^7 dt \). This gives:
   \[
   \int t^7 \ln t \, dt = \frac{t^8 \ln t}{8} - \frac{t^8}{64}.
   \]

2. For \( \int 4t^3 \, dt \), we simply get:
   \[
   \int 4t^3 \, dt = t^4.
   \]

Thus, the solution becomes:

\[
t^5 x = \frac{t^8 \ln t}{8} - \frac{t^8}{64} + t^4 + C,
\]

where \( C \) is the constant of integration.

### Step 5: Solve for \( x(t) \)
Now, solve for \( x(t) \) by dividing through by \( t^5 \):

\[
x(t) = \frac{t^3 \ln t}{8} - \frac{t^3}{64} + \frac{1}{t} + \frac{C}{t^5}.
\]

### Step 6: Apply the initial condition
Use the initial condition \( x(1) = 0 \) to find \( C \):

\[
0 = \frac{1^3 \ln 1}{8} - \frac{1^3}{64} + \frac{1}{1} + \frac{C}{1^5}.
\]

Since \( \ln 1 = 0 \), this simplifies to:

\[
0 = -\frac{1}{64} + 1 + C.
\]

Thus,

\[
C = -\frac{63}{64}.
\]

### Final Solution:
Substitute \( C = -\frac{63}{64} \) into the expression for \( x(t) \):

\[
x(t) = \frac{t^3 \ln t}{8} - \frac{t^3}{64} + \frac{1}{t} - \frac{63}{64 t^5}.
\]

This is the solution to the initial value problem.

Would you like further details on any step or have any questions?

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Here are 5 related questions:
1. What is an integrating factor, and why is it useful in solving linear differential equations?
2. How does integration by parts work, and why was it needed for \( \int t^7 \ln t \, dt \)?
3. Can you explain the significance of initial conditions in solving differential equations?
4. How does the choice of the integrating factor simplify solving the equation?
5. What other techniques can be used to solve first-order linear differential equations?

**Tip**: Always double-check initial conditions when solving differential equations to ensure the constants are correctly computed.

Solve the initial value problem. t^2 dx/dt + 5tx = t^4 ln(t) + 4, x(1) = 0.

Explore the step-by-step solution to the initial value problem involving the differential equation t^2 dx/dt + 5tx = t^4 ln(t) + 4, where x(1) = 0. This problem highlights the use of integrating factors and integration by parts to solve non-standard differential equations.

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