Question

2xy - 9x^2 + ( 2y + x^2 +1)dy/dx=0

Answer 1

Answer:

dy/dx=(3x^2-2xy)/(2y+x^2+1)

Answer 2

SOLUTION

TO EVALUATE

$\displaystyle \sf{2xy - 9 {x}^{2} + (2y + {x}^{2} + 1) \frac{dy}{dx} = 0 }$

EVALUATION

$\displaystyle \sf{2xy - 9 {x}^{2} + (2y + {x}^{2} + 1) \frac{dy}{dx} = 0 }$

$\displaystyle \sf{ \implies \: (2xy - 9 {x}^{2} )dx+ (2y + {x}^{2} + 1) dy = 0 }$

$\displaystyle \sf{ \implies \: (2xy dx + {x}^{2}dy) - 9 {x}^{2} dx+ (2y + 1) dy = 0 }$

$\displaystyle \sf{ \implies \: d( {x}^{2}y) - 9 {x}^{2} dx+ (2y + 1) dy = 0 }$

On integration we get

$\displaystyle \sf{ \implies \int \: d( {x}^{2}y) - \int \: 9 {x}^{2} dx+ \int (2y + 1) dy = c }$

$\displaystyle \sf{ \implies {x}^{2}y - \: 9 .\frac{ {x}^{3} }{3} +(2 . \frac{ {y}^{2} }{2} + y) = c }$

$\displaystyle \sf{ \implies {x}^{2}y - \: 3 {x}^{3} + {y}^{2} + y = c }$

Where C is integration constant

Hence the required solution is

$\boxed{ \displaystyle \sf{ \: \: {x}^{2}y - \: 3 {x}^{3} + {y}^{2} + y = c } \: \: }$

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