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Popular >

xy^'+y=7xy^2,y(1)=5

Solution

x y^{^{'}} + y = 7 x y^{2}, y (1) = 5

Solution

y = \frac{5}{x ( - 35 ln ( x ) + 1 )}

Solution steps

x y^{'} (x) + y = 7 x y^{2}

Solve Bernoulli ODE:

y = \frac{5}{x ( - 35 ln ( x ) + 1 )}

y = \frac{5}{x ( - 35 ln ( x ) + 1 )}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for xy^'+y=7xy^2,y(1)=5 ?
The solution for xy^'+y=7xy^2,y(1)=5 is y= 5/(x(-35ln(x)+1))