[next] [prev] [prev-tail] [tail] [up]
−x(y(x)−x)x2+y(x)2+xy′(x)−y(x)=0 ✓ Mathematica : cpu = 0.128875 (sec), leaf count = 81
{{y(x)→x(−2e2c1+x22+e2(2c1+x2)−1)2e2c1+x22+e2(2c1+x2)−1}}
✓ Maple : cpu = 0.247 (sec), leaf count = 49
{ln(2x(2(y(x))2+2x2+y(x)+x)y(x)−x)+2x22−ln(x)−_C1=0}
xy′=x(y−x)y2−x2+y
Let y=xu, then y′=u+xu′ and the above becomes
x(u+xu′)=x(xu−x)(xu)2−x2+xu(u+xu′)=(xu−x)(xu)2−x2+uxu′=(xu−x)xu2−1u′=x(u−1)u2−1
Separable.
du(u−1)u2−1=xdx−u−1u2−1=x22+C
But y=xu, hence
−yx−1(yx)2−1=x22+C
Let yx=z
−z−1z2−1=x22+C−z−1=z2−1(x22+C)(−z−1)2=(z2−1)(x22+C)2z2+1+2z=z2(x22+C)2−(x22+C)2z2(1−(x22+C)2)+2z+1+(x22+C)2=0
Solving for z (quadratic formula, some conditions apply), one of the solutions is
z=4Cx2+4C2+x4+44Cx2+4C2+x4−4
Hence
y=x4Cx2+4C2+x4+44Cx2+4C2+x4−4
Need to work on verification. Kamke gives the final solution as
y=x−2Cx2+C2+x4+4−2Cx2+C2+x4−4
I am not sure where my error now is. Need to look at this again.
[next] [prev] [prev-tail] [front] [up]