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 y2 - z2 = 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