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# what is the general solution of a differential equation of the form dy/dx = C - Ky^4?

where C and K are constants

### 2 Answers

- hfshawLv 71 decade agoFavorite Answer
This is a separable equation:

dy/dx = c - k*y^4 = -k*(y^4 - c/k)

dy/(y^4 - c/k) = - k dx

For clarity in what follows, let a^4 = c/k.

Then:

dy/(y^4 - a^4) = -k dx

Expand the left hand side in terms of partial fractions:

(1/(4a^3))*[1/(y-a) - 1/(y+a) - 2a/(y^2 + a^2) = -k dx

Integrate:

(1/(4a^3))*[ln(y-a) - ln(y+a) - 2*arctan(y/a) = -k*x + c

where c is the constant of integration.

ln((y-a)/(y+a)) - 2*arctan(y/a) = 4*(a^3)*(c - k*x)

Backsubstituting for a:

ln((y - (c/k)^(1/4))/(y + (c/k)^(1/4))) - 2*arctan(y/(c/k)^(1/4)) = 4*((c/k)^(3/4))*((c/k)^(1/4) - k*x)

This is an implicit solution for y, and I I think this is as good as you are going to get. There does not appear to be a closed-form solution for y in terms of elementary functions.