(dy)/(dx) = (x+1)/(2-y) , y ne 2...

`(dy)/(dx) = (x+1)/(2-y) , y ne 2`

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`x^(2) + y^(2) + 2x-4y+c=0`

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निम्नलिखित अवकल समीकरण को हल करें| (dy)/(dx) = (e^(x)+1)y
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(dy)/(dx) = (x+1)/(2-y) , y ne 2
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अवकल समीकरण sec^(2)x tan y dx+ sec^(2) y tan x dy= 0 का हल है
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ydx + xdy = xy(dy -dx)
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(1-x^(2))dy + xy dx = xy^(2)dx
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tan y dx+ sec^(2)y tan x dy=0
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" cosec " x logy (dy)/(dx) +x^(2)y=0
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(dy)/(dx) = (1+x^(2))(1+y^(2))
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2 sin 3y dx + 3x cos 3ydy=0
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(dy)/(d) = x/(x^(2)+1)
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(e^(x) + e^(-x)) (dy)/(dx) = (e^(x)-e^(-x))
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(x+2) (dy)/(dx) =x^(2) + 4x-9
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(dy)/(dx) = sin^(3)x cos^(2)x + xe^(x)
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(dy)/(dx) = (1- cosx)/(1+ cosx)
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y log y dx - x dy=0
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(dy)/(dx) + y=1, y ne 1
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(x+1) (dy)/(dx) =2xy
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x^(5) (dy)/(dx)= -y^(5)
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