# Answer in Differential Equations

Given DE is

“6y^2dx-x(2x^3+y)dy=0”

After rearranging the above equation ,we can write

“frac{dx}{dy}-frac{x}{6y}=frac{x^4}{3y^2}\nimplies -frac{3frac{dx}{dy}}{x^4}+frac{1}{2yx^3}=-frac{1}{y^2}”

Put

“v=frac{1}{x^3}implies frac{dv}{dy}= -frac{3frac{dx}{dy}}{x^4}”

Thus, we get

“frac{dv}{dy}+frac{1}{2y}v=-frac{1}{y^2}”

Hence, Integrating Factor is

“I.F=e^{int 1/2ydy}=sqrt{y}”

So it follows

“frac{d}{dy}(sqrt{y}v)=-y^{-3/2}\nimplies sqrt{y}v=-int y^{-3/2}dy=2y^{-1/2}+C_1\nimplies v=C_1y^{-1/2}+2y^{-1}\nimplies frac{1}{x^3}=C_1y^{-1/2}+2y^{-1}\”

After simplification we get,

“y{left(x right)} = frac{3 C_{1} sqrt{x^{9} left(9 C_{1}^{2} x^{3} + 8right)}}{2} + frac{x^{3} left(9 C_{1}^{2} x^{3} + 4right)}{2}\ny{left(x right)} = – frac{3 C_{1} sqrt{x^{9} left(9 C_{1}^{2} x^{3} + 8right)}}{2} + frac{x^{3} left(9 C_{1}^{2} x^{3} + 4right)}{2}”

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