a.) solve the differential equation using power series methods: (1-2x)y\'-2y=0 b.) find the unknown constant using the initial condition: y(0)=3/2 Solution (1-2x) dy/dx=2y dy/dx=2y/1-2x dy/y= 2 dx/1-2x integrating on both sides we get ln y =2 (-2) ln(1-2x)+c given y(0)=3/2 substitute x=0 in the eqn we get ln(y(0)) = -4 *0+c ln (3/2)=c therefore ln y= 4 ln(1/1-2x)+ln(3/2).

1. a.) solve the differential equation using power series methods: (1-2x)y'-2y=0 b.) find the
unknown constant using the initial condition: y(0)=3/2
Solution
(1-2x) dy/dx=2y dy/dx=2y/1-2x dy/y= 2 dx/1-2x integrating on both sides we get ln
y =2 (-2) ln(1-2x)+c given y(0)=3/2 substitute x=0 in the eqn we get ln(y(0)) = -4 *0+c ln
(3/2)=c therefore ln y= 4 ln(1/1-2x)+ln(3/2)