Question

What is the general solution of the differential equation ? `2xdy/dx = 10x^3y^5+y`

Answer 1

`y = (sqrt(x))/root(4)((C - 4x^5))`

Explanation:

We have:

`2xdy/dx = 10x^3y^5+y`

We can rewrite as:

`dy/dx = 5x^2y^5+y/(2x)`
`:. dy/dx - y/(2x) = 5x^2y^5`

This is a Bernoulli equitation which has a standard method to solve. Let:

`u = y^(-4) => (du)/dy = -4y^(-5)` and `dy/(du) = -y^5/4`

By the chain rule we have;

`dy/dx = dy/(du)*(du)/dx`

Substituting into the last DE we get;

`(-y^5/4)((du)/dx) -y/(2x) = 5x^2y^5`
`:. (du)/dx + 2y^(-4)/(x) = -20x^2`
`:. (du)/dx + 2/xu = -20x^2`

So the substitution has reduced the DE into a first order linear differential equation of the form:

`(d zeta)/dx + P(x) zeta = Q(x)`

We solve this using an Integrating Factor

`I = exp( \ int \ P(x) \ dx )`
`\ \ = exp( int \ 2/x \ dx )`
`\ \ = exp( 2 lnx )`
`\ \ = exp( lnx^2 )`
`\ \ = x^2`

And if we multiply the last by this Integrating Factor, `I`, we will have a perfect product differential;

`:. (du)/dx + 2/xu = -20x^2`
`:. x^2(du)/dx + 2xu = -20x^4`
`:. d/dx (x^2u) = -20x^4`

Which is now a trivial separable DE, so we can "separate the variables" to get:

`x^2u = int \ -20x^4 \ dx`

And integrating gives us:

`\ \ \ \ \ x^2u = -20x^5/5 + C`
`:. x^2u = -4x^5 + C`

Restoring the substitution we get:

`\ x^2y^(-4) = -4x^5 + C`
`:. y^(-4) = (C - 4x^5)/x^2`
`:. \ \ \ y^4 = x^2/(C - 4x^5)`
`:. \ \ \ \ \ y = (x^2/(C - 4x^5))^(1/4)`
`:. \ \ \ \ \ y = (sqrt(x))/root(4)((C - 4x^5))`