What is the solution for xy^'+y=x^4y^3 ?

The solution for xy^'+y=x^4y^3 is y= 1/(xsqrt(-x^2+c_{1))},y=-1/(xsqrt(-x^2+c_{1))}

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xy^'+y=x^4y^3

x y^{^{'}} + y = x^{4} y^{3}

y = \frac{1}{x - x ^{2} + c _{1}}, y = - \frac{1}{x - x ^{2} + c _{1}}

Solution steps

x y^{'} (x) + y = x^{4} y^{3}

Solve Bernoulli ODE:

y = \frac{1}{x - x ^{2} + c _{1}}, y = - \frac{1}{x - x ^{2} + c _{1}}

y = \frac{1}{x - x ^{2} + c _{1}}, y = - \frac{1}{x - x ^{2} + c _{1}}

\frac{d}{d x} ((6 x - 7)^{3} (8 x^{2} + 9)^{2})

\int_{0}^{5} (5 - x) d x

x y d y = y d x

\frac{4}{x}, \frac{4}{x ^{2}}, x = 7

9 y^{^{''}} - y = x e^{\frac{x}{3}}

What is the solution for xy^'+y=x^4y^3 ?
The solution for xy^'+y=x^4y^3 is y= 1/(xsqrt(-x^2+c_{1))},y=-1/(xsqrt(-x^2+c_{1))}