Math Problem Statement

Answer it using Variable separable

(x^3 + 2) y dx = x (y^4 + 3) dy at x=1, y=1. Solve using variable separation.

Solution

The given differential equation in the image is:

(x3+2)ydx=x(y4+3)dy(x^3 + 2) y \, dx = x (y^4 + 3) \, dy

To solve it using the method of separation of variables, let's proceed step by step:

Step 1: Rearrange the equation

Separate the variables xx and yy to opposite sides.

(x3+2)xdx=(y4+3)ydy\frac{(x^3 + 2)}{x} \, dx = \frac{(y^4 + 3)}{y} \, dy

Step 2: Simplify both sides

Simplify the expressions on both sides:

x3+2xdx=y4+3ydy\frac{x^3 + 2}{x} \, dx = \frac{y^4 + 3}{y} \, dy

This becomes:

(x2+2x)dx=(y3+3y)dy(x^2 + \frac{2}{x}) \, dx = (y^3 + \frac{3}{y}) \, dy

Step 3: Integrate both sides

Now, integrate both sides with respect to their respective variables.

  • For the left side, we need to integrate x2+2xx^2 + \frac{2}{x}:

(x2+2x)dx=x33+2lnx+C1\int (x^2 + \frac{2}{x}) \, dx = \frac{x^3}{3} + 2 \ln |x| + C_1

  • For the right side, we integrate y3+3yy^3 + \frac{3}{y}:

(y3+3y)dy=y44+3lny+C2\int (y^3 + \frac{3}{y}) \, dy = \frac{y^4}{4} + 3 \ln |y| + C_2

Step 4: Combine results

The general solution becomes:

x33+2lnx=y44+3lny+C\frac{x^3}{3} + 2 \ln |x| = \frac{y^4}{4} + 3 \ln |y| + C

Where C=C2C1C = C_2 - C_1 is a constant.

Step 5: Use initial condition

We are given the initial condition x=1x = 1, y=1y = 1. Substitute these into the equation to solve for CC:

133+2ln1=144+3ln1+C\frac{1^3}{3} + 2 \ln |1| = \frac{1^4}{4} + 3 \ln |1| + C

This simplifies to:

13+0=14+0+C\frac{1}{3} + 0 = \frac{1}{4} + 0 + C

C=1314=412312=112C = \frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}

Final solution:

x33+2lnx=y44+3lny+112\frac{x^3}{3} + 2 \ln |x| = \frac{y^4}{4} + 3 \ln |y| + \frac{1}{12}

This is the solution to the given differential equation.

Would you like more details or clarification on any of the steps?

5 Related Questions:

  1. How do you determine when to use variable separation in differential equations?
  2. What other methods exist to solve differential equations besides separation of variables?
  3. How can we use exact equations to solve this differential equation?
  4. What role do logarithms play in solving differential equations?
  5. Can initial conditions always be used to solve for the constant of integration?

Tip:

Always check if an equation can be rearranged to separate variables before trying more complex methods like integrating factors or exact equations.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Separation of Variables
Initial Conditions

Formulas

(x^3 + 2) y dx = x (y^4 + 3) dy
Separation of variables: dy/dx = f(x)g(y)
Integration: ∫f(x)dx = ∫g(y)dy

Theorems

Separation of Variables
Initial Value Problem

Suitable Grade Level

Undergraduate