Complex Math Simple Life

I Order Differential Equations Part I

Posted in differential, equation by beauangel on the February 16, 2007

Standard Form

\frac{dy}{dx} = f(x).g(x)

To solve:

  • Make sure that the given differential equation is in the standard form of a variable separable equation.
  • Collect the function of x and the differential dx to one side and the function of y and dy to the other side of the differential equation.
  • Integrate both sides with respect to the corresponding variables and get the solution. Add the constant of integration to one side.

An example

Find the solution of the differential equation

\frac{dy}{dx}= \frac{(x^2+1\sqrt{y^3+1}}{xy^2}, given that y^3=\tfrac{5}{4} when x = e.

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

\therefore \int (y^3+1)^{- \tfrac{1}{2}}.y^2 dy = \int \frac{(x^2+1)}{x}dx

\frac{1}{3} \frac{(y^3+1)^{\tfrac{1}{2}}}{1/2} = \frac{x^2}{2} + \ln x + c

\therefore the general solution:

\frac{2}{3}\sqrt{y^3+1} = \frac{1}{2}x^2+\ln x+c

Given: when x=e, y^3 = \tfrac{5}{4}

Substituting, we get \frac{2}{3}\sqrt{\frac{5}{4}+1} = \frac{1}{2}e^2 + \ln e +c

\frac{2}{3}\sqrt{\frac{9}{4}} = \frac{1}{2}e^2 + 1 + c

\left( \frac{2}{3}\right) \left( \frac{3}{2}\right) = \frac{1}{2}e^2 + 1 + c

1 = \frac{1}{2}e^2 + 1 + c

\therefore c = -\frac{1}{2}e^2

the particular solution:

\frac{2}{3}\sqrt{y^3+1} = \frac{1}{2}x^2 + \ln x - \frac{1}{2}e^2

ie, \frac{2}{3}\sqrt{y^3+1} = \frac{1}{2}(x^2-e^2) + \ln x.

Leave a Comment

Leave a Reply