To solve the given differential equation with the initial condition, we follow a structured approach. The differential equation is:

$x \ln(x) = y \left(1 + \sqrt{3 + y^2}\right) y'$

with the initial condition:

$y(1) = 1$

Step 1: Simplify and Rearrange the Equation

First, let's rewrite the differential equation to make it easier to handle:

$x \ln(x) = y \left(1 + \sqrt{3 + y^2}\right) \frac{dy}{dx}$

This can be rewritten as:

$\frac{dy}{dx} = \frac{x \ln(x)}{y \left(1 + \sqrt{3 + y^2}\right)}$

Step 2: Separate Variables

Separate the variables $x$ and $y$:

$\left(1 + \sqrt{3 + y^2}\right) y \, dy = x \ln(x) \, dx$

Step 3: Integrate Both Sides

Integrate both sides to find the solution.

Integrating the Left Side:

$\int \left(1 + \sqrt{3 + y^2}\right) y \, dy$

Let's break it into two parts:

$\int y \, dy + \int y \sqrt{3 + y^2} \, dy$

For the first integral:

$\int y \, dy = \frac{y^2}{2}$

For the second integral, use the substitution $u = 3 + y^2$:

$du = 2y \, dy$

$\int y \sqrt{3 + y^2} \, dy = \frac{1}{2} \int \sqrt{u} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{1}{3} (3 + y^2)^{3/2}$

So the left side integral is:

$\frac{y^2}{2} + \frac{1}{3} (3 + y^2)^{3/2}$

Integrating the Right Side:

$\int x \ln(x) \, dx$

Use integration by parts, where $u = \ln(x)$ and $dv = x \, dx$:

$\int x \ln(x) \, dx = x^2 \frac{\ln(x)}{2} - \int x \frac{1}{2} \, dx = x^2 \frac{\ln(x)}{2} - \frac{x^2}{4}$

So the right side integral is:

$\frac{x^2 \ln(x)}{2} - \frac{x^2}{4}$

Step 4: Combine the Results

Combine the integrated results:

$\frac{y^2}{2} + \frac{1}{3} (3 + y^2)^{3/2} = \frac{x^2 \ln(x)}{2} - \frac{x^2}{4} + C$

Use the initial condition $y(1) = 1$ to find the constant $C$:

Step 5: Apply the Initial Condition

Substitute $x = 1$ and $y = 1$:

$\frac{1^2}{2} + \frac{1}{3} (3 + 1^2)^{3/2} = \frac{1^2 \ln(1)}{2} - \frac{1^2}{4} + C$

$\frac{1}{2} + \frac{1}{3} \cdot 4^{3/2} = 0 - \frac{1}{4} + C$

$\frac{1}{2} + \frac{1}{3} \cdot 8 = -\frac{1}{4} + C$

$\frac{1}{2} + \frac{8}{3} = -\frac{1}{4} + C$

$\frac{3}{6} + \frac{16}{6} = -\frac{1}{4} + C$

$\frac{19}{6} = -\frac{1}{4} + C$

$C = \frac{19}{6} + \frac{1}{4} = \frac{38}{12} + \frac{3}{12} = \frac{41}{12}$

Step 6: Write the Final Solution

$\frac{y^2}{2} + \frac{1}{3} (3 + y^2)^{3/2} = \frac{x^2 \ln(x)}{2} - \frac{x^2}{4} + \frac{41}{12}$

This is the implicit solution to the given differential equation satisfying the initial condition $y(1) = 1$.

Would you like more details or have any questions?

Relative Questions

1. What is the method of separation of variables in differential equations?
2. How do you perform integration by parts?
3. What are common techniques for solving first-order differential equations?
4. How do you apply initial conditions to find particular solutions?
5. What are substitution methods in solving integrals?
6. How do you verify solutions to differential equations?
7. What are the applications of differential equations in real-life scenarios?
8. How do numerical methods help in solving complex differential equations?

Tip: When solving differential equations, carefully check each step of integration and substitution, as small errors can lead to incorrect solutions.