Do not know how to handle complex numbers

msd1997

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Apr 22, 2021
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How do i simplify the value of c2?
Is this correct so far?
 

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nasi112

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When you get a complex number the solution will have a sine and cosine functions

like this

\(\displaystyle f(x) = c_1 e^x \cos x + c_2 e^x \sin x\)

\(\displaystyle e^x\) will take the real number and \(\displaystyle \cos x\) and \(\displaystyle \sin x\) will take the imaginary number

you have \(\displaystyle m = 1 + 3i\) and \(\displaystyle m = 1 - 3i\)

this means you have \(\displaystyle 1\) as real number and \(\displaystyle 3i\) as an imaginary number

so the solution will be

\(\displaystyle f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)

Now, you can continue with the conditions to find \(\displaystyle c_1\) and \(\displaystyle c_2\)
 

msd1997

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Apr 22, 2021
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When you get a complex number the solution will have a sine and cosine functions

like this

\(\displaystyle f(x) = c_1 e^x \cos x + c_2 e^x \sin x\)

\(\displaystyle e^x\) will take the real number and \(\displaystyle \cos x\) and \(\displaystyle \sin x\) will take the imaginary number

you have \(\displaystyle m = 1 + 3i\) and \(\displaystyle m = 1 - 3i\)

this means you have \(\displaystyle 1\) as real number and \(\displaystyle 3i\) as an imaginary number

so the solution will be

\(\displaystyle f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)

Now, you can continue with the conditions to find \(\displaystyle c_1\) and \(\displaystyle c_2\)
I have one doubt.
One of the roots is m= 1-3i which becomes sin3x. So do we ignore the negative value? What if it was 1+ 4i ? Would it been sin4x ?
 

msd1997

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Joined
Apr 22, 2021
Messages
19
When you get a complex number the solution will have a sine and cosine functions

like this

\(\displaystyle f(x) = c_1 e^x \cos x + c_2 e^x \sin x\)

\(\displaystyle e^x\) will take the real number and \(\displaystyle \cos x\) and \(\displaystyle \sin x\) will take the imaginary number

you have \(\displaystyle m = 1 + 3i\) and \(\displaystyle m = 1 - 3i\)

this means you have \(\displaystyle 1\) as real number and \(\displaystyle 3i\) as an imaginary number

so the solution will be

\(\displaystyle f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)

Now, you can continue with the conditions to find \(\displaystyle c_1\) and \(\displaystyle c_2\)
Also, what will be Y(0) ? I know e^x becomes 1. But what happens to cos3x and sin3x ?
 

msd1997

New member
Joined
Apr 22, 2021
Messages
19
When you get a complex number the solution will have a sine and cosine functions

like this

\(\displaystyle f(x) = c_1 e^x \cos x + c_2 e^x \sin x\)

\(\displaystyle e^x\) will take the real number and \(\displaystyle \cos x\) and \(\displaystyle \sin x\) will take the imaginary number

you have \(\displaystyle m = 1 + 3i\) and \(\displaystyle m = 1 - 3i\)

this means you have \(\displaystyle 1\) as real number and \(\displaystyle 3i\) as an imaginary number

so the solution will be

\(\displaystyle f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)

Now, you can continue with the conditions to find \(\displaystyle c_1\) and \(\displaystyle c_2\)


Sorry... The ignoring negative sign was a stupid question... I have figured that out myself.
 

Subhotosh Khan

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Also, what will be Y(0) ? I know e^x becomes 1. But what happens to cos3x and sin3x ?
cos(3*0) = cos(0) =1 and

sin(3*0) = sin(0) = 0
 

nasi112

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Aug 23, 2020
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417
I have one doubt.
One of the roots is m= 1-3i which becomes sin3x. So do we ignore the negative value? What if it was 1+ 4i ? Would it been sin4x ?
You can always ignore the negative value and treat \(\displaystyle -3i\) and \(\displaystyle 3i\) as just \(\displaystyle 3i\)

Why?

because \(\displaystyle 1+3i\), will give you this equation
\(\displaystyle y_1(x) = e^x \cos 3x + e^x \sin 3x \)

and

\(\displaystyle 1-3i\), will give you this equation
\(\displaystyle y_2(x) = e^x \cos -3x + e^x \sin -3x = e^x \cos 3x - e^x \sin 3x\)

the solution is their combination

\(\displaystyle u(x) = y_1(x) + y_2(x) = 2e^x \cos 3x \)
\(\displaystyle u(x) = \frac{1}{2}y_1(x) + \frac{1}{2}y_2(x) = e^x \cos 3x \)

we can have another solution as
\(\displaystyle v(x) = y_1(x) - y_2(x) = 2ie^x \sin 3x \)
\(\displaystyle v(x) = \frac{1}{2i}y_1(x) - \frac{1}{2i}y_2(x) = e^x \sin 3x \)

the general solution is just a combination of the two

\(\displaystyle y(x) = c_1 u(x) + c_2 v(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)
 

msd1997

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Apr 22, 2021
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You can always ignore the negative value and treat \(\displaystyle -3i\) and \(\displaystyle 3i\) as just \(\displaystyle 3i\)

Why?

because \(\displaystyle 1+3i\), will give you this equation
\(\displaystyle y_1(x) = e^x \cos 3x + e^x \sin 3x \)

and

\(\displaystyle 1-3i\), will give you this equation
\(\displaystyle y_2(x) = e^x \cos -3x + e^x \sin -3x = e^x \cos 3x - e^x \sin 3x\)

the solution is their combination

\(\displaystyle u(x) = y_1(x) + y_2(x) = 2e^x \cos 3x \)
\(\displaystyle u(x) = \frac{1}{2}y_1(x) + \frac{1}{2}y_2(x) = e^x \cos 3x \)

we can have another solution as
\(\displaystyle v(x) = y_1(x) - y_2(x) = 2ie^x \sin 3x \)
\(\displaystyle v(x) = \frac{1}{2i}y_1(x) - \frac{1}{2i}y_2(x) = e^x \sin 3x \)

the general solution is just a combination of the two

\(\displaystyle y(x) = c_1 u(x) + c_2 v(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)
Can you please tell me if my solution is correct?
 

nasi112

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Aug 23, 2020
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417
Your derivative is wrong

the derivative of

\(\displaystyle f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)

Is Not

\(\displaystyle f'(x) = c_1 e^x(-3\sin 3x) + c_2 e^x (3\cos 3x)\)

Did you forget the product rule when you have two functions multiplying by each other such as \(\displaystyle e^x \cos 3x\)?
 

msd1997

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Apr 22, 2021
Messages
19
Your derivative is wrong

the derivative of

\(\displaystyle f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)

Is Not

\(\displaystyle f'(x) = c_1 e^x(-3\sin 3x) + c_2 e^x (3\cos 3x)\)

Did you forget the product rule when you have two functions multiplying by each other such as \(\displaystyle e^x \cos 3x\)?

I'm sorry. I'm a beginner. Can you please tell me if my solution is correct , or have I made further mistakes?
 

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nasi112

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Aug 23, 2020
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I'm sorry. I'm a beginner. Can you please tell me if my solution is correct , or have I made further mistakes?
Yup, you got it right. Congratulations.

:)
No worries if you're a beginner. We all were once beginners.
 
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