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Popular >

y^'=1-y,y(0)=0

Solution

y^{^{'}} = 1 - y, y (0) = 0

Solution

y = - e^{- t} + 1

Solution steps

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

Solve separable ODE:

y = - e^{- t} + 1

y = - e^{- t} + 1

Popular Examples

xy(dx)/(dy)=y+2

x y \frac{d x}{d y} = y + 2

y^'=(cos(x)e^{sin(x)})/(2y)

y^{^{'}} = \frac{cos ( x ) e ^{s i n (x)}}{2 y}

(dP)/(dt)=5sqrt(Pt),P(1)=5

\frac{d P}{d t} = 5 Pt, P (1) = 5

cos(y)(2ln(x)-x^3)(dy)/(dx)= 2/x-3x^2

cos (y) (2 ln (x) - x^{3}) \frac{d y}{d x} = \frac{2}{x} - 3 x^{2}

(dy}{dx}y=-arccot(\frac{6x)/3)

\frac{d y}{d x} y = - arccot (\frac{6 x}{3})

Frequently Asked Questions (FAQ)

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