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Popular >

-y^'+(4y)/x =x^3y^2

Solution

- y^{^{'}} + \frac{4 y}{x} = x^{3} y^{2}

Solution

y = \frac{8 x ^{4}}{x ^{8} + c _{1}}

Solution steps

- y^{'} (x) + \frac{4 y}{x} = x^{3} y^{2}

Solve Bernoulli ODE:

y = \frac{8 x ^{4}}{x ^{8} + c _{1}}

y = \frac{8 x ^{4}}{x ^{8} + c _{1}}

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x \to 2 lim (x^{2} - 2 x + 4)

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\int cos (y x) d x

Frequently Asked Questions (FAQ)

What is the solution for -y^'+(4y)/x =x^3y^2 ?
The solution for -y^'+(4y)/x =x^3y^2 is y=(8x^4)/(x^8+c_{1)}