Question

Solve: $y'=\frac{6 x^{2}-5xy-2 y^{2}}{6x^{2}-8xy+y^{2}}$

Answer 1

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Answer 2

Answer:

$6x^{2} y-8xy^{2} +\frac{y^{3} }{3}$ = $2x^{3} - 5y\frac{x^{2} }{2} -2y^{2}x$

Step-by-step explanation:

y'=$\frac{6x^{2}-5xy-2y^{2} }{6x^{2}-8xy+y^{2} }$

since,

$\frac{dy}{dx} = \frac{6x^{2}-5xy-2y^{2}}{6x^{2}-8xy+y^{2} }$

So,

$({6x^{2}-5xy-2y^{2} }) \, dx = ({6x^{2}-8xy+y^{2} }) \, dy$

integrating both side we get ,

$\int({6x^{2}-5xy-2y^{2} }) \, dx = \int({6x^{2}-8xy+y^{2} }) \, dy$

After integrating both sides  we get ,

$6\frac{x^{3} }{3} - 5y\frac{x^{2} }{2} -2y^{2}x$=$6x^{2} y-8xy^{2} +\frac{y^{3} }{3}$

 $2x^{3} - 5y\frac{x^{2} }{2} -2y^{2}x$ = $6x^{2} y-8xy^{2} +\frac{y^{3} }{3}$