How do you solve this differential equation?

Question

9u"+12u'+4u=0, u(0)=a>0, u'(0)=-1

solve the initial vaue problem and find the critical value of a that seperates solutions that become negative from those that are always positive.

Answer 1

This is a simple, homogeneous equation with constant coefficients. I'm guessing that u' = du/dt and
u'' = d²u/du² so solve the auxillary equation
9x² + 12x + 4 = 0 which has a repeated root x = -2/3 and so u = c_1*e^(-2t/3) + c_2*t*e^(-2t/3)
Now go back and take the derivative to get the coefficients c_1 and c_2. In this case, at t = 0, c_1 = a (from the problemn statement) so that a < 0 => y is negative for some time.


Doug

Answer 2

The corresponding characteristic equation is
9r^2+12r+4=0
Which gives (3r+2)^2=0
So the solution is r=-2/3 and the solutions are u=e^(-2/3 t) and u=t e^(-2/3 t). You should be able to handle the rest.

Answer 3

http://math.uww.edu/~mcfarlat/250de2.htm

http://hyperphysics.phy-astr.gsu.edu/hbase/diff.html