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Popular >

y^'+by=0

Solution

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

Solution

y = e^{- b t + c_{1}}

Solution steps

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

Solve separable ODE:

y = e^{- b t + c_{1}}

y = e^{- b t + c_{1}}

Popular Examples

integral of (x^2)/(sqrt((2-x^2)))

\int \frac{x ^{2}}{( 2 - x ^{2} )} d x

integral of sinh(10x)

\int sinh (10 x) d x

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

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

derivative of 1/(t^2)derivative of

\frac{1}{t ^{2}}

integral from 0 to 1 of x*e^x

\int_{0}^{1} x \cdot e^{x} d x

Frequently Asked Questions (FAQ)

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