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Popular >

y^'=g-y

Solution

y^{^{'}} = g - y

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Graph

Popular Examples

y^'=(4x+7)(5y+4)

y^{^{'}} = (4 x + 7) (5 y + 4)

x^{^{'}} = x^{2} (t)

\frac{d v}{d x} + xv = 0

(dv)/(dt)v=(2^{5/2})t^{3/2}+3

\frac{d v}{d t} v = (2^{\frac{5}{2}}) t^{\frac{3}{2}} + 3

(dy)/(dx)(sin(pi)x+cos(pi)y)^2=2

\frac{d y}{d x} (sin (π) x + cos (π) y)^{2} = 2

Frequently Asked Questions (FAQ)

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