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Popular >

y^'+ab^2y=0

Solution

y^{^{'}} + a b^{2} y = 0

Solution

y = e^{- a b^{2} t + c_{1}}

Solution steps

y^{'} (t) + a b^{2} y = 0

Solve separable ODE:

y = e^{- a b^{2} t + c_{1}}

y = e^{- a b^{2} t + c_{1}}

Popular Examples

x u^{^{'}} - a H (x) = 0

t^{^{'}} = (2 t - 1)^{3}

(dy)/(dx)=((3x^2-e^x))/(2y-5),y(0)=1

\frac{d y}{d x} = \frac{( 3 x ^{2} - e ^{x} )}{2 y - 5}, y (0) = 1

(dy)/(dx)=10-y/(50)

\frac{d y}{d x} = 10 - \frac{y}{50}

(dy)/(dx)=5x^4y^2

\frac{d y}{d x} = 5 x^{4} y^{2}

Frequently Asked Questions (FAQ)

What is the solution for y^'+ab^2y=0 ?
The solution for y^'+ab^2y=0 is y=e^{-ab^2t+c_{1}}