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  • Find the general solution. (x + 3 y^2) dy/dx = y (y greater than 0)

Find the general solution.

    Q12.    (x+3y^2)\frac{dy}{dx} = y\ (y > 0)

Answers (1)

best_answer

Given equation is
(x+3y^2)\frac{dy}{dx} = y\ (y > 0)
we can rewrite it as
\frac{dx}{dy}-\frac{x}{y}= 3y
This is  \frac{dx}{dy} + px = Q  type where p =\frac{-1}{y} and Q =3y
Now,
I.F. = e^{\int pdy}= e^{\int \frac{-1}{y} dy}= e^{-\log y } =y^{-1}= \frac{1}{y}                     
Now, the solution of given differential equation is given by relation
x(I.F.) =\int (Q\times I.F.)dy +C
x(\frac{1}{y}) =\int 3y\times \frac{1}{y}dy +C
\frac{x}{y}= \int 3dy + C
\frac{x}{y}= 3y+ C
x = 3y^2+Cy
Therefore, the general solution is   x = 3y^2+Cy

Posted by

Gautam harsolia

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