The general solution of the differential equa | vedclass

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Question

(A) Given differential equation is $xy(y+2) dy + (y^3-1) dx = 0$. Rearranging the terms,we get $\frac{dx}{x} + \frac{y(y+2)}{y^3-1} dy = 0$. Integrati

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$\log |x| + \frac{1}{3} \log |y^3-1| + \frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{2y+1}{\sqrt{3}} ight) = c$

Vedclass · Answer

(A) Given differential equation is $xy(y+2) dy + (y^3-1) dx = 0$. Rearranging the terms,we get $\frac{dx}{x} + \frac{y(y+2)}{y^3-1} dy = 0$. Integrating both sides,$\int \frac{dx}{x} + \int \frac{y^2+2y}{y^3-1} dy = c$. Using partial fractions for $\frac{y^2+2y}{(y-1)(y^2+y+1)} = \frac{A}{y-1} + \frac{By+C}{y^2+y+1}$. Solving for constants,we find $A = 1$,$B = 0$,$C = 1$. So,$\int \frac{dx}{x} + \int \frac{1}{y-1} dy + \int \frac{1}{y^2+y+1} dy = c$. $\log |x| + \log |y-1| + \int \frac{1}{(y+1/2)^2 + (\sqrt{3}/2)^2} dy = c$. $\log |x(y-1)| + \frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{2y+1}{\sqrt{3}} ight) = c$. Since $y^3-1 = (y-1)(y^2+y+1)$,the solution is $\log |x| + \frac{1}{3} \log |y^3-1| + \frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{2y+1}{\sqrt{3}} ight) = c$.

The general solution of the differential equation $x y(y+2) dy + (y^3-1) dx = 0$ is

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