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Popular >

y^'+y-1=0

Solution

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

Solution

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

Solution steps

y^{'} (t) + y - 1 = 0

Solve separable ODE:

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

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

Popular Examples

7 y y^{^{'}} = x, y (7) = 3

y^'-x/(5y^4)=0,y(0)=2,y(2)=a

y^{^{'}} - \frac{x}{5 y ^{4}} = 0, y (0) = 2, y (2) = a

(dy)/(dx)=sqrt(y^2-25)

\frac{d y}{d x} = y^{2} - 25

(1+x^2)dy-sqrt(1-y^2)dx=0

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

(dy)/(dx)=14xsec(y)

\frac{d y}{d x} = 14 x sec (y)

Frequently Asked Questions (FAQ)

What is the solution for y^'+y-1=0 ?
The solution for y^'+y-1=0 is y=e^{-t-c_{1}}+1