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Popular >

y^'=(1+y)/(x-1)

Solution

y^{^{'}} = \frac{1 + y}{x - 1}

Solution

y = e^{c_{1}} x - e^{c_{1}} - 1

Solution steps

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

Solve separable ODE:

y = e^{c_{1}} x - e^{c_{1}} - 1

y = e^{c_{1}} x - e^{c_{1}} - 1

Popular Examples

y^{^{'}} = \frac{1}{y ^{1}}

(dy)/(dx)=6(5-y)

\frac{d y}{d x} = 6 (5 - y)

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

\frac{d y}{d x} = \frac{2 y ( 2 x - 1 ) e ^{x}}{2 x + 1}

y^'=1+4y^2,y(0)=0

y^{^{'}} = 1 + 4 y^{2}, y (0) = 0

y^'=((e^{-x}-e^x))/((3+4y))

y^{^{'}} = \frac{( e ^{- x} - e ^{x} )}{( 3 + 4 y )}

Frequently Asked Questions (FAQ)

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