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anuksha0456

The solution to the homogenous differential equation, (11y² - xy)dx + x²dy = 0, is [tex]y = \frac{x}{11 ln (x) + C}[/tex].

A homogeneous differential equation has a homogeneous function as one of its components, if f(λx, λy) = λⁿf(x, y), for any non-zero constant λ, the function is said to be homogenous. The homogeneous differential equation's generic form is of the type f(x, y).dy + g(x, y).dx = 0. The degree of the homogeneous differential equation is the same for the equation's variables x and y.

In the question, we are asked to solve the homogeneous differential equation, (11y² - xy)dx + x²dy = 0.

The given equation can be solved as follows:

  • Grouping by differentials, gives us: x²dy = (xy - 11y²)dx.
  • We now substitute, u = yx, making y = ux, or, dy = u.dx + x.du.
  • Making the substitutions, we get: x²(u.dx + x.du) = (u - 11u²)x²dx.
  • Expanding the parentheses, we get: ux².dx + x³.du = ux².dx - 11u²x².dx.
  • Reducing ux².dx, we get: x³.du = -11u²x².dx.
  • Dividing by x³ and u², we get du/u² = -11/x.
  • Now, we integrate both sides of the equation: [tex]\int \frac{1}{u^2}du = \int -\frac{11}{x}dx[/tex]
  • Calculating the resulting integrals, we get: -1/u = C - 11 ln(x).
  • Undoing the substitution, u = y/x, we get: -x/y = C - 11 ln(x)
  • The final solution is: [tex]y = \frac{x}{11 ln (x) + C}[/tex]

Thus, the solution to the homogenous differential equation, (11y² - xy)dx + x²dy = 0, is [tex]y = \frac{x}{11 ln (x) + C}[/tex].

The provided question is incomplete. The complete question is:

"Use the method for solving homogenous equations to solve the following differential equation.

(11y² - xy)dx + x²dy = 0".

Learn more about solving homogenous differential equation at

https://brainly.com/question/13512633

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