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Popular >

xy^'+y=4xy^2,y(1)=9

Solution

x y^{^{'}} + y = 4 x y^{2}, y (1) = 9

Solution

y = \frac{9}{x ( - 36 ln ( x ) + 1 )}

Solution steps

x y^{'} (x) + y = 4 x y^{2}

Solve Bernoulli ODE:

y = \frac{9}{x ( - 36 ln ( x ) + 1 )}

y = \frac{9}{x ( - 36 ln ( x ) + 1 )}

Popular Examples

integral of-8e^{x+4}-5/(x+7)-5sec^2(x+6)

\int - 8 e^{x + 4} - \frac{5}{x + 7} - 5 sec^{2} (x + 6) d x

limit as x approaches-1+of x+4

x \to - 1 + lim (x + 4)

(dx)/(dy)=xysin(y^2)

\frac{d x}{d y} = x y sin (y^{2})

derivative of (2e^{-1/(x^2})/(x^3))

\frac{d}{d x} (\frac{2 e ^{- \frac{1}{x ^{2}}}}{x ^{3}})

(\partial)/(\partial z)(4z)

\frac{\partial}{\partial z} (4 z)

Frequently Asked Questions (FAQ)

What is the solution for xy^'+y=4xy^2,y(1)=9 ?
The solution for xy^'+y=4xy^2,y(1)=9 is y= 9/(x(-36ln(x)+1))