Question

(2xy cosx^2-2xy+1)dx+(sinx^2-x^2)dy=0

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

SOLUTION

TO DETERMINE

The solution is

$\sf{(2xy \cos {x}^{2} - 2xy + 1)dx + ( \sin {x}^{2} - {x}^{2} )dy = 0}$

EVALUATION

$\sf{(2xy \cos {x}^{2} - 2xy + 1)dx + ( \sin {x}^{2} - {x}^{2} )dy = 0}$

$\sf{ \implies \: (2xy \cos {x}^{2}dx + \sin {x}^{2}dy) -( 2xy dx +{x}^{2} dy ) + 1.dx = 0}$

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

On integration we get

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

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

Where C is integration constant

FINAL ANSWER

Hence the required solution is

$\sf{ y\sin {x}^{2} - {x}^{2} y + x = c}$

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