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Please help in solving this ordinary diffrential equation: y" + y=0. I thank you for your help.

Guest Dec 30, 2015

Best Answer 

 #2
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+10

Solve ( d^2 y(x))/( dx^2)+y(x) = 0:
Assume a solution will be proportional to e^(lambda x) for some constant lambda.
Substitute y(x)  =  e^(lambda x) into the differential equation:
( d^2 )/( dx^2)(e^(lambda x))+e^(lambda x)  =  0
Substitute ( d^2 )/( dx^2)(e^(lambda x))  =  lambda^2 e^(lambda x):
lambda^2 e^(lambda x)+e^(lambda x)  =  0
Factor out e^(lambda x):
(lambda^2+1) e^(lambda x)  =  0
Since e^(lambda x) !=0 for any finite lambda, the zeros must come from the polynomial:
lambda^2+1  =  0
Solve for lambda:
lambda = i or lambda = -i
The roots lambda  =  ± i give y_1(x) = c_1 e^(i x), y_2(x) = c_2 e^(-i x) as solutions, where c_1 and c_2 are arbitrary constants.
The general solution is the sum of the above solutions:
y(x)  =  y_1(x)+y_2(x)  =  c_1 e^(i x)+c_2/e^(i x)
Apply Euler's identity e^(alpha+i beta) = e^alpha cos(beta)+i e^alpha sin(beta):
y(x)  =  c_1 (cos(x)+i sin(x))+c_2 (cos(x)-i sin(x))
Regroup terms:
y(x)  =  (c_1+c_2) cos(x)+i (c_1-c_2) sin(x)
Redefine c_1+c_2 as c_1 and i (c_1-c_2) as c_2, since these are arbitrary constants:
Answer: | y(x)  =  c_1 cos(x)+c_2 sin(x)

Guest Dec 30, 2015
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4+0 Answers

 #1
avatar+91001 
+5

I started by thinking it had sin or cos in it and I narrowed it to y=asin(ix)

Where a is any real number.

 

I think that is right be maybe there are lots of answers.

 

Here is another one

Y = ke^(ix+b)

Where k and b are any

real numbers.

Maybe they could be complex numbers too ? Not sure....  

Melody  Dec 30, 2015
 #2
avatar
+10
Best Answer

Solve ( d^2 y(x))/( dx^2)+y(x) = 0:
Assume a solution will be proportional to e^(lambda x) for some constant lambda.
Substitute y(x)  =  e^(lambda x) into the differential equation:
( d^2 )/( dx^2)(e^(lambda x))+e^(lambda x)  =  0
Substitute ( d^2 )/( dx^2)(e^(lambda x))  =  lambda^2 e^(lambda x):
lambda^2 e^(lambda x)+e^(lambda x)  =  0
Factor out e^(lambda x):
(lambda^2+1) e^(lambda x)  =  0
Since e^(lambda x) !=0 for any finite lambda, the zeros must come from the polynomial:
lambda^2+1  =  0
Solve for lambda:
lambda = i or lambda = -i
The roots lambda  =  ± i give y_1(x) = c_1 e^(i x), y_2(x) = c_2 e^(-i x) as solutions, where c_1 and c_2 are arbitrary constants.
The general solution is the sum of the above solutions:
y(x)  =  y_1(x)+y_2(x)  =  c_1 e^(i x)+c_2/e^(i x)
Apply Euler's identity e^(alpha+i beta) = e^alpha cos(beta)+i e^alpha sin(beta):
y(x)  =  c_1 (cos(x)+i sin(x))+c_2 (cos(x)-i sin(x))
Regroup terms:
y(x)  =  (c_1+c_2) cos(x)+i (c_1-c_2) sin(x)
Redefine c_1+c_2 as c_1 and i (c_1-c_2) as c_2, since these are arbitrary constants:
Answer: | y(x)  =  c_1 cos(x)+c_2 sin(x)

Guest Dec 30, 2015
 #3
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0

Thank you very much, Guest.

Guest Dec 30, 2015
 #4
avatar
+5

Tha Auxiliary Equation is

\(\displaystyle m^{2} + 1 = 0\) ,

which has roots

\(\displaystyle \pm \imath\) ,

which means that the general solution is

\(\displaystyle y = Ce^{\imath x}+De^{-\imath x}\)

which is more conveniently written as

\(\displaystyle y = A\cos x+B \sin x\) .

Guest Dec 30, 2015

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