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Popular >

ay^'+by=c

Solution

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

Solution

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

Solution steps

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

Graph

Popular Examples

(dx)/(dt)=18-(9x)/(10)

\frac{d x}{d t} = 18 - \frac{9 x}{10}

u^'=((\H(t))/t)^{1/3}

u^{^{'}} = (\frac{H ( t )}{t})^{\frac{1}{3}}

sqrt(4x^2+2x^2)dx-4y*dy=8x*dx

4 x^{2} + 2 x^{2} d x - 4 y \cdot d y = 8 x \cdot d x

(ln(y))^5(dy)/(dx)=x^5y,y(1)=e^2

(ln (y))^{5} \frac{d y}{d x} = x^{5} y, y (1) = e^{2}

y^'=(1-11x)y^2,y(0)=-1/9

y^{^{'}} = (1 - 11 x) y^{2}, y (0) = - \frac{1}{9}

Frequently Asked Questions (FAQ)

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