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Popular >

(1-x)y^'=y

Solution

(1 - x) y^{^{'}} = y

Solution

y = \frac{c _{1}}{1 - x}

Solution steps

(1 - x) y^{'} (x) = y

Solve separable ODE:

y = \frac{c _{1}}{1 - x}

y = \frac{c _{1}}{1 - x}

Popular Examples

y^{^{'}} = 6 y^{2} x

y \cdot y^{^{'}} = - x

(1+x)y^'=e^{-y}-1

(1 + x) y^{^{'}} = e^{- y} - 1

(-3(2x-u)+6x+2)dx-(u-1)(2dx-du)=0

(- 3 (2 x - u) + 6 x + 2) d x - (u - 1) (2 d x - d u) = 0

(e^x+1)^2e^{-y}dx+(e^x+1)^3e^{-x}dy=0

(e^{x} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{3} e^{- x} d y = 0

Frequently Asked Questions (FAQ)

What is the solution for (1-x)y^'=y ?
The solution for (1-x)y^'=y is y=(c_{1})/(1-x)