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

use the analogous quadratic equation
4r^2 - 4r + 1 = 0

The only solution is r=0.5. Therefore the solution is
y = a*exp(x/2) + b*x*exp(x/2). you would need to know some values of y to determine a and b

Answer 3

4y''-4y'+y=0

The variable "x" is missing, so set v=y'

Then y''=dv/dx=(dv/dy)(dy/dx)=v(dv/dy)

We have

4v(dv/dy)-4v+y=0

4v(dv/dy)-4v=-y

Now we have the general form of an ordinary, first order differential equation.

Integrating factor: e^int(-4dy)=e^(-4y)

Now integrate -ye^(-4y)dy by parts, then solve.

General solution:

y=[int(-ye^(-4y)dy)+C]/e^(-4y)

where int=integral.