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Popular >

ay^'+y=0

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

limit as x approaches-infinity of 3e^x

x \to - \infty lim (3 e^{x})

derivative of x^2+i

\frac{d}{d x} (x^{2} + i)

integral of (1+ln(x)+y/x)

\int (1 + ln (x) + \frac{y}{x}) d x

derivative of f(x)=e^7+ln(6)derivative of

f (x) = e^{7} + ln (6)

simplify 1/((z-1))*1/(e^{z+1)}simplify

\frac{1}{( z - 1 )} \cdot \frac{1}{e ^{z + 1}}

Frequently Asked Questions (FAQ)

What is the solution for ay^'+y=0 ?
The solution for ay^'+y=0 is y=e^{(-t+c_{1})/a}