Solve the differential equation $$ \frac{d x}{d t}=\frac{x \ln x}{t} $$ Assume $$ x, t0 $$, and use the initial condition $$ x(1)=5 $$ $$ x= $$

Question

Solve the differential equation $$ \frac{d x}{d t}=\frac{x \ln x}{t} $$ Assume $$ x, t>0 $$, and use the initial condition $$ x(1)=5 $$ $$ x= $$

UpStudy AI · Accepted Answer

We start with the differential equation
$$
\frac{dx}{dt} = \frac{x \ln x}{t}.
$$
Since $$ x>0 $$ and $$ t>0 $$, we can separate variables by rewriting the equation as
$$
\frac{dx}{x \ln x} = \frac{dt}{t}.
$$

1. **Integrate both sides:**  
   $$
   \int \frac{dx}{x \ln x} = \int \frac{dt}{t}.
   $$

2. **Solve the left integral by substitution:**  
   Let
   $$
   u = \ln x \quad \text{so that} \quad du = \frac{dx}{x}.
   $$
   Then, the integral becomes
   $$
   \int \frac{du}{u} = \ln |u| + C_1 = \ln |\ln x| + C_1.
   $$

3. **Solve the right integral:**  
   $$
   \int \frac{dt}{t} = \ln |t| + C_2.
   $$

4. **Combine the results:**  
   Equate the two integrals (we can combine constants into a single constant $$ C $$):
   $$
   \ln |\ln x| = \ln |t| + C.
   $$
   Since $$ x > 0 $$ and given $$ x(1)=5>1 $$, we have $$ \ln x >0 $$, so we can drop the absolute values:
   $$
   \ln (\ln x) = \ln t + C.
   $$

5. **Exponentiate both sides to solve for $$\ln x$$:**  
   $$
   \ln x = e^{\ln t + C} = e^C \cdot t.
   $$
   Denote $$ K = e^C $$ (where $$ K > 0 $$), then
   $$
   \ln x = Kt.
   $$

6. **Determine the constant $$ K $$ using the initial condition $$ x(1)=5 $$:**  
   Substitute $$ t = 1 $$ and $$ x = 5 $$ into the equation:
   $$
   \ln 5 = K \cdot 1 \quad \Longrightarrow \quad K = \ln 5.
   $$
   Therefore, the equation becomes
   $$
   \ln x = (\ln 5)t.
   $$

7. **Solve for $$ x $$:**  
   Exponentiate both sides:
   $$
   x = e^{(\ln 5)t} = \left(e^{\ln 5}\right)^t = 5^t.
   $$

Thus, the solution to the differential equation is
$$
x(t)=5^t.
$$
The solution to the differential equation is $$ x(t) = 5^t $$.

Solve the differential equation

Assume , and use the initial condition

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