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Popular >

ay^'+by=0

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

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4 t x \frac{d x}{d t} = x^{2} + 1

(\partial)/(\partial z)(cos(kz-wt))

\frac{\partial}{\partial z} (cos (k z - wt))

inverse oflaplace 1/(2s^2+s+1)inverse laplace

\frac{1}{2 s ^{2} + s + 1}

limit as x approaches 1 of (3x-4)^2

x \to 1 lim ((3 x - 4)^{2})

derivative of f(x)=5-x^2 derivative of

f (x) = 5 - x^{2}

Frequently Asked Questions (FAQ)

What is the solution for ay^'+by=0 ?
The solution for ay^'+by=0 is y=e^{(-bt+c_{1})/a}