Answer on Differential Equations Question for noor
( x^2 + cosy ) + ( y^2+x+sinx) dy/dx = x
We can transform our equation to this form :
( x^2 + cosy )dx + ( y^2+x+sinx )dy = xdx
( x^2 - x + cosy )dx + ( y^2+x+sinx )dy = 0 .
In this equation I(x,y) = x^2 - x + cosy and J(x,y) = y^2+x+sinx .
Then we find : dI/dy = -siny and dJ/dx = 1 + cosx.
We see that dI/dy not equal dJ/dx .
That's why given equation is not exact.
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