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Popular >

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

Solution

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

Solution

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

Solution steps

y = (x + 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

e^yy^'=((ln(t))^2)/t ,y(1)=ln(4)

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (4)

(y^2-1)dx+xdy=0

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

y^3(dy)/(dx)=3x^4y^4-12x^4

y^{3} \frac{d y}{d x} = 3 x^{4} y^{4} - 12 x^{4}

y^{^{'}} - y \cdot tan (x) = 0

\frac{d y}{d t} = - 2 t y

Frequently Asked Questions (FAQ)

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