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What is a general solution to the differential equation #y'=5x^(2/3)y^4#?

Answer 1

Ezra Adams

# y = root(3)(1/(C -9x^(5/3))) #

# dy/dx = 5x^(2/3)y^4 #

This is a First Order separable DE, so collecting like terms and "separating the variables" we get;

# int y^-4dy = int5x^(2/3) dx #

Integrating we get:

# y^-3/-3 = 5x^(5/3)/(5/3) + C_1 # # y^-3 = -9x^(5/3) - 3C_1 # # 1/y^3 = -9x^(5/3) + C # # y^3 = 1/(C -9x^(5/3)) # # y = root(3)(1/(C -9x^(5/3))) #

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Answer 2

Charles Anctil

The general solution to the differential equation ( y' = 5x^{2/3} y^4 ) is given by ( y(x) = \left( \frac{1}{c - \frac{5}{5x^{1/3}}} \right)^{1/3} ), where ( c ) is an arbitrary constant.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.