Is there an explicit ODE solution for this system?

Find conditions a,b, and c such that the system has no solution, one solution, and infinitely many solutions.?

• Answer:

  x-by=-1, x= by -1 x+ay=3, (by-1) +ay =3 by+ay= 4 y= 4/(a+b) no solution when (a+b)=0 or a=-b 2x+y-z=a 2y+3z=b x-z=c, x= z+c 2x+y-z = 2(z+c) +y-z = a 2z +2c +y -z =a z +y = a-2c y= (a-2c) -z 2y+3z=b 2(a-2c-z) -z =b 2a -4c -2z -z=b -3z = b -2a +4c z= - (b-2a+4c) / 3 x-z =c x- - (b-2a+4c) / 3 = b x+ (b-2a+4c) / 3 =b x = 3b/ (b-2a+4c) 2x+y-z=a 2[3b/ (b-2a+4c)] +y + (b-2a+4c) / 3 =a y= a - 2[3b/ (b-2a+4c)] - (b-2a+4c) / 3 you need to simplify each fraction then evaluate its nominator and denominator a fraction has -no solution where the denominator =0 - many solutions where the nominator=0 -one solution when top=bottom