Math Problem Statement

Please explain how they got yhe answer

Please explain how they got the answer for the differential equation e^(x^2) * y * y' + x = 0 with y(0) = 2.

Solution

The problem asks for the solution of the differential equation:

ex2yy+x=0withy(0)=2.e^{x^2} y y' + x = 0 \quad \text{with} \quad y(0) = 2.

Step-by-step Explanation:

  1. Rewrite the differential equation: Start by simplifying the given equation: ex2yy+x=0.e^{x^2} y y' + x = 0. Rearrange it: ex2yy=x.e^{x^2} y y' = -x. Now, divide both sides by ex2ye^{x^2} y to isolate yy': y=xex2y.y' = -\frac{x}{e^{x^2} y}.

  2. Separate variables: The next step is to use separation of variables. Multiply both sides by yy and dxdx to separate yy and xx: ydy=xex2dx.y \, dy = -x e^{-x^2} \, dx.

  3. Integrate both sides: Now, integrate both sides of the equation. The left-hand side is straightforward: ydy=12y2.\int y \, dy = \frac{1}{2} y^2. For the right-hand side: xex2dx.\int -x e^{-x^2} \, dx. Using the substitution u=x2u = x^2, with du=2xdxdu = 2x \, dx, the integral becomes: 12eudu=12ex2+C.\frac{1}{2} \int e^{-u} \, du = \frac{1}{2} e^{-x^2} + C. So, after integration, we get: 12y2=12ex2+C.\frac{1}{2} y^2 = \frac{1}{2} e^{-x^2} + C.

  4. Solve for y2y^2: Multiply both sides by 2 to simplify: y2=ex2+2C.y^2 = e^{-x^2} + 2C.

  5. Use the initial condition y(0)=2y(0) = 2: To find the constant CC, use the initial condition y(0)=2y(0) = 2. Substitute x=0x = 0 and y=2y = 2 into the equation: 22=e(0)2+2C,2^2 = e^{-(0)^2} + 2C, 4=1+2C.4 = 1 + 2C. Solving for CC: 2C=3C=32.2C = 3 \quad \Rightarrow \quad C = \frac{3}{2}.

  6. Final solution: Substitute the value of CC back into the equation: y2=ex2+3.y^2 = e^{-x^2} + 3.

This is the final solution to the differential equation.

Would you like more details on any step or do you have any questions?

Related Questions:

  1. What substitution was used to integrate the right-hand side of the equation?
  2. Why is separation of variables a useful technique for solving this type of differential equation?
  3. How would the solution change if the initial condition was different, e.g., y(0)=1y(0) = 1?
  4. Can this equation be solved using any other methods, such as an integrating factor?
  5. What is the importance of the initial condition in determining the constant CC?

Tip:

When integrating a function that includes an exponential and a polynomial, substitution often simplifies the process effectively.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differential Equations
Separation of Variables
Initial Value Problem

Formulas

y * dy = -x * e^(-x^2) * dx
∫ y dy = ∫ -x * e^(-x^2) dx
Substitution method: u = x^2

Theorems

Separation of Variables Technique
Initial Condition Application

Suitable Grade Level

Undergraduate level (Calculus II or Differential Equations course)