Solve each of following DE subject to given conditions, if any.
(
3
0
 ﻿points, Separable DE
)
 
(
l
n
y
)
y
'
=
-
x
2
y
,
y
(
0
)
=
e
.
Choose the right answer from the following possible answers:
a
.
1
2
l
n
(
y
)
=
-
1
3
x
3
+
C
b
.
1
2
(
l
n
(
y
)
)
2
=
-
1
3
x
3
+
1
2
c
.
l
n
(
y
2
)
=
x
3
+
1
2
d
.
None of the above


We are given the following differential equation:

\[
(\ln y) y' = -x^2 y, \quad y(0) = e.
\]

### Step 1: Rearranging the equation
We first rewrite the given equation in a more manageable form:

\[
(\ln y) \frac{dy}{dx} = -x^2 y.
\]

To solve this separable differential equation, we separate variables and integrate.

### Step 2: Separation of variables
First, divide both sides by \( y \):

\[
\frac{\ln y}{y} \frac{dy}{dx} = -x^2.
\]

Rearranging to separate \( y \) and \( x \):

\[
\frac{\ln y}{y} dy = -x^2 dx.
\]

### Step 3: Integration
We now integrate both sides.

- The left-hand side requires the integration of \( \frac{\ln y}{y} \), which can be done by substitution. Let \( u = \ln y \), so \( du = \frac{1}{y} dy \). The integral becomes:

\[
\int \frac{\ln y}{y} dy = \int u \, du = \frac{u^2}{2} = \frac{(\ln y)^2}{2}.
\]

- The right-hand side is straightforward:

\[
\int -x^2 dx = -\frac{x^3}{3}.
\]

Thus, after integrating both sides, we get:

\[
\frac{(\ln y)^2}{2} = -\frac{x^3}{3} + C.
\]

### Step 4: Solve for \( y \)
To solve for \( y \), multiply both sides by 2:

\[
(\ln y)^2 = -\frac{2x^3}{3} + 2C.
\]

Let \( 2C = C_1 \) (since it's an arbitrary constant, it can be renamed). So we have:

\[
(\ln y)^2 = -\frac{2x^3}{3} + C_1.
\]

### Step 5: Apply the initial condition \( y(0) = e \)
Substitute \( x = 0 \) and \( y = e \) into the equation to find \( C_1 \):

\[
(\ln e)^2 = 0 = -\frac{2(0)^3}{3} + C_1.
\]

So, \( C_1 = 0 \). Therefore, the solution simplifies to:

\[
(\ln y)^2 = -\frac{2x^3}{3}.
\]

### Step 6: Final solution and matching with options
Thus, the final form of the solution is:

\[
\frac{(\ln y)^2}{2} = -\frac{x^3}{3} + \frac{1}{2}.
\]

Comparing with the given options, the correct choice is:

**Option (b)**:
\[
\frac{(\ln y)^2}{2} = -\frac{x^3}{3} + \frac{1}{2}.
\]

### Final answer:
The correct answer is **(b)**.

---

Would you like more details or have any further questions? Here are some related problems to deepen your understanding:

1. Solve \( y' = x^2y^2 \) using separation of variables.
2. Solve the separable DE \( y' = y/x \), \( y(1) = 2 \).
3. Find the general solution of \( y' = y \ln y \).
4. Solve \( x^2y' = y + x^3 \), with \( y(1) = 0 \).
5. Verify the uniqueness of the solution for the initial condition \( y(0) = e^2 \) in this problem.

**Tip:** When solving separable differential equations, always ensure that you handle constants carefully during integration, and don’t forget to apply initial conditions if given!

Solve each of the following DE subject to the given conditions, if any. (3 points, Separable DE) (ln y)y' = -x^2y, y(0) = e. Choose the right answer from the following possible answers: a. 1/2 ln(y) = -1/3 x^3 + C b. 1/2 (ln(y))^2 = -1/3 x^3 + 1/2 c. ln(y^2) = x^3 + 1/2 d. None of the above.

This page explains how to solve the separable differential equation (ln y)y' = -x^2y, with the initial condition y(0) = e. The problem involves logarithmic functions and separation of variables. Explore the step-by-step solution to understand how to integrate both sides and apply the initial condition to find the correct answer.

Math Problem Statement

( l n y ) y '

x 2 y , y ( 0 )

e . Choose the right answer from the following possible answers: a . 1 2 l n ( y )

1 3 x 3 + C b . 1 2 ( l n ( y ) ) 2

1 3 x 3 + 1 2 c . l n ( y 2 )

Solution

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