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Popular >

ay^'+by=k

Solution

a y^{^{'}} + b y = k

Solution

y = - \frac{e ^{\frac{- b t - c _{1} b}{a}}}{b} + \frac{k}{b}

Solution steps

a y^{'} (t) + b y (t) = k

Solve separable ODE:

y = - \frac{e ^{\frac{- b t - c _{1} b}{a}}}{b} + \frac{k}{b}

y = - \frac{e ^{\frac{- b t - c _{1} b}{a}}}{b} + \frac{k}{b}

Popular Examples

(dy)/(dx)=-sqrt(y/x)

\frac{d y}{d x} = - \frac{y}{x}

y^'-y=2,te^{2t}y(0)=1

y^{^{'}} - y = 2, t e^{2 t} y (0) = 1

y d y - 2 x y^{2} d x = 0

t^2(1+y^2)+2y(dy)/(dt)=0,y(0)=1

t^{2} (1 + y^{2}) + 2 y \frac{d y}{d t} = 0, y (0) = 1

x(y^2+10)dx=y(x^2+20)dy

x (y^{2} + 10) d x = y (x^{2} + 20) d y

Frequently Asked Questions (FAQ)

What is the solution for ay^'+by=k ?
The solution for ay^'+by=k is y=-(e^{(-bt-c_{1}b)/a})/(b)+k/b