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यदि y= y(x) अवकल समीकरण x(dy)/(dx) + 2y ...

यदि `y= y(x)` अवकल समीकरण `x(dy)/(dx) + 2y = x^(2)` का हल है जो y(1) = 1, को संतुष्ट करता है , तो `y((1)/(2))` बराबर है

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यदि x^(y)=y^(x) [जहाँ x ne y ] का हल है -

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

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

Solve: (dy) / (dx) = (y (x + 2y)) / (x (2x + y)), y (1) = 2

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

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

If y^(2) -2x^(2) = y , then dy/dx at (1,-1) is

y(1+x^(2))dy=x(1+y^(2))dx

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