k is real. Solve the differential equations y' = z(y + z)^{k}, z' = y(y + z)^{k} subject to y(0) = 1, z(0) = 0.

**Solution**

Adding, (y + z)' = (y + z)^{k+1}. Integrating (y + z)^{k} = 1/(1 - kx) (1).

Multiplying opposite sides of the two given equations together, we get yy'(y + z)^{k} = zz' (y + z)^{k}. Hence yy' = zz'. Integrating y^{2} - z^{2} = 1. Hence (y + z)^{k}(y - z)^{k} = 1, so (y - z)^{k} = 1 - kx (2).

For k non-zero, we can immediately solve (1) and (2) to get y = 1/2 ( (1 - kx)^{1/k} + 1/(1 - kx)^{1/k} ), z = 1/2 ( (1 - kx)^{1/k} - 1/(1 - kx)^{1/k} ).

For k = 0, the original equations simplify to y' = z, z'= y. So y'' = y. So y = A cosh x + B sinh x, z = A cosh x + B sinh x. Applying the initial conditions, y = cosh x, z = sinh x.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002