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Popular >

ay^'+yb=c

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

(dy)/(dx)=(x+1)/(y^2),y(1)=0

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

y^'csc(t)=(-y^{21})/(20)

y^{^{'}} csc (t) = \frac{- y ^{21}}{20}

(dp)/(dt)=0.0017p(1-p/(6200))

\frac{d p}{d t} = 0.0017 p (1 - \frac{p}{6200})

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

(dy)/(dx)+y^2cos(x)=0

\frac{d y}{d x} + y^{2} cos (x) = 0

Frequently Asked Questions (FAQ)

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