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Popular >

y=1+y^'

Solution

y = 1 + y^{^{'}}

Solution

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

Solution steps

y = 1 + y^{'} (t)

Solve separable ODE:

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

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

Popular Examples

(dP)/(dt)=P(1-P/(1800))

\frac{d P}{d t} = P (1 - \frac{P}{1800})

sec (x) y^{^{'}} + y = 2

(dy)/(dx)=(1-y)*cos(x)

\frac{d y}{d x} = (1 - y) \cdot cos (x)

(2x+3)y^'=xy(2x+3)^{1/2}

(2 x + 3) y^{^{'}} = x y (2 x + 3)^{\frac{1}{2}}

sqrt(1+y^2)-xy^'=0

1 + y^{2} - x y^{^{'}} = 0

Frequently Asked Questions (FAQ)

What is the solution for y=1+y^' ?
The solution for y=1+y^' is y=-e^{t+c_{1}}+1