Initial value problem differentials, close to getting answer

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SUMMARY

The discussion centers on solving the initial value problem defined by the differential equation y" + 2y' + 10y = 0 with initial conditions y(0) = 1 and y'(0) = -1. The participant correctly applied the quadratic formula to find the roots r = 1 - 3i and r = 1 + 3i, leading to the general solution yg = c1e^t(cos 3t + i sin 3t) + c2e^t(cos 3t - i sin 3t). However, the participant encountered confusion regarding the constants c1 and c2, ultimately deriving an incorrect final solution of yg = 2e^t cos 3t instead of the expected yg = e^-t cos 3t. The error was identified as neglecting the negative sign in the quadratic formula.

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  • Understanding of second-order linear differential equations
  • Familiarity with the quadratic formula
  • Knowledge of complex numbers and Euler's formula
  • Ability to differentiate functions and apply initial conditions
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  • Study the application of initial conditions in determining constants in differential equations
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Homework Statement


y" + 2y' + 10y=0
y(0)=1
y'(0)= -1

solve initial value problem

Homework Equations



e^( ~ + iu)t= e^ ~t (cos ut + i sin ut)

The Attempt at a Solution



i've gotten pretty far into the problem, but i just can't seem to get the correct final answer.

I changed y" + 2y' + 10y=0 int r^2 +2r +10= 0 and used the quadratic formula to get:

r= 1-3i and r= 1+ 3i

so for the general solution i got:
yg= y1 + y2
yg= c1e^t( cos 3t + i sin 3t) + c2e^t( cos 3t - i sin 3t)

i need to solve for the constants:

y= c1 e^t cos 3t + c2 i sin3t ( an example in the book followed this method but i don't know why there is a c2 next to sin 3t and why there isn't a y2 )

i got c1= 1 and took the derivative to find c2 and got c2= -2/3

y'= c1 e^t cos 3t -3c1e^t sin 3t + c2e^tsin3t + 3 cos3tc2e^t

overall i got yg= 2e^tcos 3t but the answer is suppose to be yg= e^-t cos 3t. why? the textbook i use has this annoying ability of skipping 90% of the arithmatic work within examples, so I'm not sure where I could have gone wrong, can u guys help? :/
 
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itunescape said:

Homework Statement


y" + 2y' + 10y=0
y(0)=1
y'(0)= -1

solve initial value problem

Homework Equations



e^( ~ + iu)t= e^ ~t (cos ut + i sin ut)

The Attempt at a Solution



i've gotten pretty far into the problem, but i just can't seem to get the correct final answer.

I changed y" + 2y' + 10y=0 int r^2 +2r +10= 0 and used the quadratic formula to get:

r= 1-3i and r= 1+ 3i

so for the general solution i got:
yg= y1 + y2
yg= c1e^t( cos 3t + i sin 3t) + c2e^t( cos 3t - i sin 3t)

i need to solve for the constants:

y= c1 e^t cos 3t + c2 i sin3t ( an example in the book followed this method but i don't know why there is a c2 next to sin 3t and why there isn't a y2 )

i got c1= 1 and took the derivative to find c2 and got c2= -2/3

y'= c1 e^t cos 3t -3c1e^t sin 3t + c2e^tsin3t + 3 cos3tc2e^t

overall i got yg= 2e^tcos 3t but the answer is suppose to be yg= e^-t cos 3t. why? the textbook i use has this annoying ability of skipping 90% of the arithmatic work within examples, so I'm not sure where I could have gone wrong, can u guys help? :/

remember the quadratic eq'n formula is


r=\frac{-b \pm \sqrt{b^2-4ac}}{2a}


you just forgot the '-b'
 

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