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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution of the differential equation $y\;dx+(x-y^2)dy=0$

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  • The first order linear differential equation of the form $\large\frac{dx}{dy}$$+ Px = Q$ can be solved as follows:
  • (i) Write the given equation in the form of $\large\frac{dx}{dy}$$ + Px = Q$
  • (ii) Find the integrating factor (I.F) = $e^{\int Pdy}$.
  • (iii) Write the solution as y(I.F) = integration of $Q(I.F) dy + C$
Step 1:
Using the information in the tool box, first let us rewrite the given equation
dividing throughout by $ydy$ we get,
$\large\frac{dx}{dy} +\frac{ (x-y^2)}{y }$$= 0$
$\large\frac{dx}{dy} +\frac{ x}{y }$$= y$
Here $P = \large\frac{1}{y}$ and $Q = y$
Step 2:
To find the integral factor,$\int pdy = \int\large\frac{ dy}{y} =$$ \log y$
Hence $I.F = e^{\large\log y }= y$
The required solution is $x.y = \int y.ydy$
Step 3:
$\int y^2dy=\large\frac{y^3}{3}$
Hence the required solution is $xy = \large\frac{y^3}{x }$$+ C$
dividing throughout by $y$ we get
$x = \large\frac{y^2}{3 }+\frac{ C}{y}$
This is the required solution.
answered Jul 31, 2013 by sreemathi.v

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