Find the solution 
y
(
x
)
y(x) of the initial-value problem 
(
ln
⁡
x
)
d
y
d
x
+
1
x
y
=
3
x
3
(lnx) 
dx
dy
​
 + 
x
1
​
 y=3x 
3
  with 
x
>
1
x>1 and 
y
(
e
)
=
7
y(e)=7.



y
(
x
)
=
y(x)=

We are given the initial-value problem:

\[
(\ln x) \frac{dy}{dx} + \frac{1}{x} y = 3x^3
\]
with the condition \( y(e) = 7 \), and \( x > 1 \).

### Step 1: Rewrite the equation

This is a first-order linear differential equation of the form:
\[
\frac{dy}{dx} + P(x)y = Q(x)
\]
where \( P(x) = \frac{1}{x \ln x} \) and \( Q(x) = \frac{3x^3}{\ln x} \).

Now, divide the entire equation by \( \ln x \) (since \( \ln x \neq 0 \) for \( x > 1 \)):
\[
\frac{dy}{dx} + \frac{1}{x \ln x} y = \frac{3x^3}{\ln x}
\]

### Step 2: Find the integrating factor

The integrating factor \( \mu(x) \) is given by:
\[
\mu(x) = \exp\left( \int \frac{1}{x \ln x} dx \right)
\]
Let's compute this integral. Set \( u = \ln x \), so \( du = \frac{1}{x} dx \), and the integral becomes:
\[
\int \frac{1}{x \ln x} dx = \int \frac{1}{u} du = \ln |u| = \ln (\ln x)
\]
Thus, the integrating factor is:
\[
\mu(x) = \exp(\ln (\ln x)) = \ln x
\]

### Step 3: Multiply through by the integrating factor

Multiplying both sides of the differential equation by \( \ln x \):
\[
\ln x \frac{dy}{dx} + \frac{1}{x} y = 3x^3
\]

Notice that the left-hand side is the derivative of \( y \ln x \) (by the product rule):
\[
\frac{d}{dx}(y \ln x) = 3x^3
\]

### Step 4: Integrate both sides

Now integrate both sides with respect to \( x \):
\[
y \ln x = \int 3x^3 dx
\]
\[
y \ln x = \frac{3x^4}{4} + C
\]

### Step 5: Solve for \( y(x) \)

Divide by \( \ln x \) to isolate \( y(x) \):
\[
y(x) = \frac{3x^4}{4 \ln x} + \frac{C}{\ln x}
\]

### Step 6: Use the initial condition

We are given \( y(e) = 7 \). Substituting \( x = e \) and \( \ln e = 1 \) into the equation:
\[
7 = \frac{3e^4}{4} + C
\]
Solve for \( C \):
\[
7 = \frac{3e^4}{4} + C \quad \Rightarrow \quad C = 7 - \frac{3e^4}{4}
\]

### Final Solution:

The solution to the differential equation is:
\[
y(x) = \frac{3x^4}{4 \ln x} + \frac{7 - \frac{3e^4}{4}}{\ln x}
\]

Would you like more details or have any questions about the steps?

Here are 5 related questions for practice:

1. Solve a similar equation where \( y(e) = 5 \) and the differential equation is \( (\ln x) \frac{dy}{dx} + \frac{1}{x} y = 2x^3 \).
2. What is the general solution to the equation \( y' + P(x)y = Q(x) \)?
3. How would the solution change if \( y(e) = 0 \) instead?
4. Can you solve the same differential equation but for \( x > 0 \)?
5. What is the importance of the integrating factor in solving linear differential equations?

**Tip**: Always check the initial condition after solving the differential equation to ensure the solution fits.

Find the solution y(x) of the initial-value problem (lnx) dy/dx + 1/x y = 3x^3 with x > 1 and y(e) = 7.

Learn how to solve the first-order differential equation (ln x) dy/dx + (1/x)y = 3x^3 with the initial condition y(e) = 7. This step-by-step solution covers key concepts such as integrating factors and product rule. Suitable for undergraduate students studying calculus and differential equations.

Math Problem Statement

Find the solution y ( x ) y(x) of the initial-value problem ( ln ⁡ x ) d y d x + 1 x y

1 x>1 and y ( e )

y ( x )

Solution

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