differential equation?

Question

plz solve the following differential equation completely....
4D2y-4D1y+y=0;
where D1 &D2 denotes the and second derivatives respectively...

Answer 1

We normally write this as
4y" - 4y' + y = 0
This is a second-order linear constant coefficient homogeneous ODE, so it's not hard to solve.
First we write the characteristic equation:
4r^2 - 4r + 1 = 0
<=> (2r - 1) = 0
so we have a repeated solution at r = 1/2.
So the solution is
y = A e^(x/2) + Bx e^(x/2).

(If the solutions to the characteristic equation are real and distinct, say r1 and r2, then y = Ae^(r1 x) + Be^(r2 x); if they are real and equal, say r, then y = Ae^(rx) + Bx e^(rx); if they are complex conjugate, say r = a ± ib, then y = A e^(ax) cos bx + B e^(ax) sin bx.)

Answer 2

Put the equation into auxiliary form:
4m^2 - 4m + 1 =0,
(2m-1))(2m -1)=0
Roots are repeated with m=1/2
Therefore the general solution is, with constants a and b,
y = a*e^(x/2) + b*e^(x/2)