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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

(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}

(dy)/(dx)x^2+x^2y^2+y^2=0

\frac{d y}{d x} x^{2} + x^{2} y^{2} + y^{2} = 0

(dy)/(dx)=((xy-4x+y-4))/((xy-6x-3y+18))

\frac{d y}{d x} = \frac{( x y - 4 x + y - 4 )}{( x y - 6 x - 3 y + 18 )}

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}}