Diff. eq dx/dy = (xy^3) / (2y^4 + x^4) is not separable. Find the general solution. use subst. x=y*u then...?

Question

algebraiclly simplify the eq. to one with the variable (this time, y and u ) separable.

Answer 1

(dx/dy) = (xy^3) / (2y^4 + x^4)

Let x = y*u.
Then (dx/dy) = u + y*(du/dy).

Substituting these in the given equation we have,

u + y*(du/dy) = (u*y^4) / {2y^4 + (u*y)^4} = u/(2 + u^4)

or y*(du/dy) = u/(2 + u^4) - u = -u*(1 + u^4)/(2 + u^4)

or ∫ [-(2 + u^4)/{u*(1 + u^4)}] du = ∫ [1/y] dy _________(1)



Now, let us first express [-(2 + u^4)/{u*(1 + u^4)}] in easily integrable form.

-(2 + u^4)/{u*(1 + u^4)}

= -{2*(1 + u^4) - u^4}/{u*(1 + u^4)}

= -{2*(1 + u^4)/{u*(1 + u^4)} + u^4}/{u*(1 + u^4)}

= -2/u + u^3/(1 + u^4)

Therefore, ∫ [-(2 + u^4)/{u*(1 + u^4)}] du

= ∫ [ -2/u + u^3/(1 + u^4)] du

= -2∫ [1/u] du + ∫ [ u^3/(1 + u^4)] du

= -2*ln(u) + (1/4)*ln(1 + u^4) + k

where k = arbitrary constant of integration

Hence, from (1) we have

-2*ln(u) + (1/4)*ln(1 + u^4) + k = ln(y)

or (1/4)*ln(1 + u^4) + k = ln(y) + 2*ln(u) = ln(y) + ln(u^2) = ln(y*u^2)

or k = ln(y*u^2) - (1/4)*ln(1 + u^4)

= ln(y*u^2) - ln[(1 + u^4)^(1/4)]

= ln[(y*u^2)/{(1 + u^4)^(1/4)}]


Now, substituting u = x/y we have

k = ln[(x^2/y)/{(1 + (x/y)^4)^(1/4)}]

or k = ln[(x^2)/{(x^4 + y^4)^(1/4)}]

Let k = ln(c).

Then c = (x^2)/{(x^4 + y^4)^(1/4)}

or x^2 = c*(x^4 + y^4)^(1/4)

Answer 2

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