Math Problem Statement

Find a particular solution of x*y' = y + (2*x^4/y) cos(x^2), y(sqrt(pi/2)) = sqrt(pi).

Solution

To find a particular solution to the given differential equation

xy=y+(2x4y)cos(x2),xy' = y + \left(\frac{2x^4}{y}\right) \cos(x^2),

with the initial condition

y(π2)=π,y\left(\sqrt{\frac{\pi}{2}}\right) = \sqrt{\pi},

we can approach this step-by-step. Here's how to solve it:

Step 1: Simplify the Differential Equation

First, let's rewrite the differential equation in a more standard form: xyy=2x4ycos(x2).xy' - y = \frac{2x^4}{y} \cos(x^2).

Step 2: Introduce a Substitution

Given the structure of this equation, consider using a substitution to simplify. One potential substitution is y=x2f(x)y = x^2 f(x), as this might cancel some terms. Let's define: y=x2f(x).y = x^2 f(x).

Step 3: Differentiate the Substitution

Differentiate y=x2f(x)y = x^2 f(x) with respect to xx: y=2xf(x)+x2f(x).y' = 2x f(x) + x^2 f'(x).

Step 4: Substitute into the Original Differential Equation

Substitute yy and yy' into the original differential equation: x[2xf(x)+x2f(x)]=x2f(x)+2x4x2f(x)cos(x2).x[2x f(x) + x^2 f'(x)] = x^2 f(x) + \frac{2x^4}{x^2 f(x)} \cos(x^2).

Simplify this expression: 2x2f(x)+x3f(x)=x2f(x)+2x4cos(x2)x2f(x).2x^2 f(x) + x^3 f'(x) = x^2 f(x) + \frac{2x^4 \cos(x^2)}{x^2 f(x)}.

2x2f(x)+x3f(x)=x2f(x)+2x2cos(x2)f(x).2x^2 f(x) + x^3 f'(x) = x^2 f(x) + \frac{2x^2 \cos(x^2)}{f(x)}.

Step 5: Isolate Terms

Bring like terms together: x3f(x)=x2f(x)+2x2cos(x2)f(x),x^3 f'(x) = -x^2 f(x) + \frac{2x^2 \cos(x^2)}{f(x)},

which simplifies to: xf(x)=f(x)+2cos(x2)f(x).x f'(x) = -f(x) + \frac{2 \cos(x^2)}{f(x)}.

Step 6: Separation of Variables

Separate the variables if possible. This equation can be rearranged to: f(x)f(x)+2cos(x2)f(x)=1x.\frac{f'(x)}{-f(x) + \frac{2 \cos(x^2)}{f(x)}} = \frac{1}{x}.

This is a complicated expression, so solving this exactly may require more advanced techniques or numerical methods.

Step 7: Apply Initial Condition

Now, use the initial condition y(π2)=πy\left(\sqrt{\frac{\pi}{2}}\right) = \sqrt{\pi} to find the particular solution: π22f(π2)=πf(π2)=π.\sqrt{\frac{\pi}{2}}^2 f\left(\sqrt{\frac{\pi}{2}}\right) = \pi f\left(\sqrt{\frac{\pi}{2}}\right) = \sqrt{\pi}.

Thus, f(π2)=1π.f\left(\sqrt{\frac{\pi}{2}}\right) = \frac{1}{\sqrt{\pi}}.

Final Step: Particular Solution

Given the complexity of this differential equation, finding a closed-form solution may not be straightforward. It may require numerical methods to solve completely. However, this is the setup and how you would approach finding the particular solution.

Would you like further steps on numerical methods for this differential equation, or any clarification on the process so far?

Further Questions

  1. How would you approach solving this differential equation using numerical methods?
  2. Can you explain the choice of substitution and its impact on simplifying the equation?
  3. What other substitutions could potentially simplify this differential equation?
  4. What role does the initial condition play in finding the particular solution?
  5. How can we check if a function is a solution to a differential equation?

Tip

When dealing with complex differential equations, it's often helpful to look for substitutions or transformations that simplify the terms. Numerical methods like Euler's or Runge-Kutta can be useful when analytic solutions are hard to find.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Calculus
Trigonometry

Formulas

xy' - y = (2x^4/y) cos(x^2)
y = x^2 f(x)
y' = 2x f(x) + x^2 f'(x)

Theorems

Method of substitution for differential equations
Separation of variables

Suitable Grade Level

University