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Popular >

ax^'+bx=0

Solution

a x^{^{'}} + b x = 0

Solution

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

Solution steps

a x^{'} (t) + b x (t) = 0

Solve separable ODE:

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

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

Popular Examples

xy-2y-x+2-(xy-2x-y+2)(dy)/(dx)=0

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

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(x^{2} + x - 6) y + (x + 3) y^{^{'}} + (x - 2) y = 0

y - 3 y^{^{'}} - 18 = 0

(dy)/(dt)y=(ln(6t))^2

\frac{d y}{d t} y = (ln (6 t))^{2}

t y^{^{'}} + y = 5

Frequently Asked Questions (FAQ)

What is the solution for ax^'+bx=0 ?
The solution for ax^'+bx=0 is x=e^{(-bt+c_{1})/a}