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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([tex]\sqrt{3}[/tex]*x/2)+bsin([tex]\sqrt{3}[/tex]*x/2))

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

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([tex]\sqrt{3}[/tex]*x/2)+bsin([tex]\sqrt{3}[/tex]*x/2))

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

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