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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution of the differential equation $(x+3y^2)\large\frac{dy}{dx}=y$$\;(y>0)$

$\begin{array}{1 1} (A)\;x = 3y +Cy\\(B)\;x = 3y -Cy \\ (C)\;x = -3y -Cy \\(D)\;x = 2y +Cy \end{array} $

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1 Answer

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Toolbox:
  • To solve first order linear differnetial equation of the form $\large\frac{dy}{dx}$$ + Py = Q$
  • (i)Write the equation in the form $\large\frac{dx}{dy}$$ + Px = Q$
  • (ii) Find the integral factor $(I.F) = e^{\int pdy}$
  • (iii) Write the required solution as $y.(I.F) = \int Q.(I.F) dy + C$
Step 1:
Using the information in the tool box, let us rewrite the equation in the form $\large\frac{dx}{dy}$$ + Px + Q$
divide on both sides by $(x+3y^2)$
$\large\frac{dy}{dx}=\frac{y}{(x+3y^2)}$
reciprocate on both sides,
$\large\frac{dx}{dy} =\frac{ (x+3y^2)}{y}$
$\large\frac{dx}{dy }-\large\frac{ x}{y}$$ = 3y$
Here $P = \large\frac{-1}{y}$ and $Q = 3y$
Step 2:
To find the integral factor let us integrate P,
$\int -\large\frac{ dy}{y }=$$- \log y = \log\big(\large\frac{1}{y}\big)$
Hence $I.F = e^{\large\log[\large\frac{1}{y}] }=\large\frac{1}{ y}$
Hence the required solution is $xy = \int 3y.(\large\frac{1}{y}) $$+ C$
Step 3:
$ \int 3 dy = 3y$
Hence the required solution is $\large\frac{x}{y} $$= 3y.y + C$
$\large\frac{x}{y}$$= 3y^2 + C$
$x = 3y +Cy$
This is the required equation.
answered Jul 31, 2013 by sreemathi.v
 

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