# Question in complex general equation (2nd order)

Question:
Find the real-valued general solution of the differential equation

y''+1y'+1y=0
where primes indicate differentiation with respect to x. (Use the parameters a, b, etc., for undetermined constants in your solution.)

My attempt:
Use characteristic equation:
r^2+r+1=0
I used the quadratic formula and got r=(-1 +-sqrt(-3))/2
So we get a=-.5 b-sqrt(3)/2

Following e^(ax)(c1cos(bx)+c2sin(bx):

we get e^(-.5x)(a*cos($$\sqrt{3}$$*x/2)+bsin($$\sqrt{3}$$*x/2))

Anyone see any flaw i nlogic of algebra or math? I can't seem to get a correct answer

Last edited:

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tiny-tim
Homework Helper
hi batmankiller! (have a square-root: √ and try using the X2 icon just above the Reply box )
Find the real-valued general solution of the differential equation

y''+1y'+1y=0
where primes indicate differentiation with respect to x. (Use the parameters a, b, etc., for undetermined constants in your solution.)

we get e^(-.5x)(a*cos($$\sqrt{3}$$*x/2)+bsin($$\sqrt{3}$$*x/2))
looks ok to me have you tried entering x/2 (instead of 0.5 x), or (√3/2)x, or 0.866 x ?

umm yeah thanks so much. -.5x apparently doesn't equal -x/2 anymore. Math as we know it has changed!

tiny-tim
the computers have taken over we have to speak the language of the conquerors! 