Do not know how to handle complex numbers

[msd1997]

How do i simplify the value of c2?
Is this correct so far?

 

[nasi112]

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

like this

[MATH]f(x) = c_1 e^x \cos x + c_2 e^x \sin x[/MATH]
[MATH]e^x[/MATH] will take the real number and [MATH]\cos x[/MATH] and [MATH]\sin x[/MATH] will take the imaginary number

you have [MATH]m = 1 + 3i[/MATH] and [MATH]m = 1 - 3i[/MATH]
this means you have [MATH]1[/MATH] as real number and [MATH]3i[/MATH] as an imaginary number

so the solution will be

[MATH]f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x[/MATH]
Now, you can continue with the conditions to find [MATH]c_1[/MATH] and [MATH]c_2[/MATH]

 

[msd1997]

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

like this

[MATH]f(x) = c_1 e^x \cos x + c_2 e^x \sin x[/MATH]
[MATH]e^x[/MATH] will take the real number and [MATH]\cos x[/MATH] and [MATH]\sin x[/MATH] will take the imaginary number

you have [MATH]m = 1 + 3i[/MATH] and [MATH]m = 1 - 3i[/MATH]
this means you have [MATH]1[/MATH] as real number and [MATH]3i[/MATH] as an imaginary number

so the solution will be

[MATH]f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x[/MATH]
Now, you can continue with the conditions to find [MATH]c_1[/MATH] and [MATH]c_2[/MATH]

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]

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

like this

[MATH]f(x) = c_1 e^x \cos x + c_2 e^x \sin x[/MATH]
[MATH]e^x[/MATH] will take the real number and [MATH]\cos x[/MATH] and [MATH]\sin x[/MATH] will take the imaginary number

you have [MATH]m = 1 + 3i[/MATH] and [MATH]m = 1 - 3i[/MATH]
this means you have [MATH]1[/MATH] as real number and [MATH]3i[/MATH] as an imaginary number

so the solution will be

[MATH]f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x[/MATH]
Now, you can continue with the conditions to find [MATH]c_1[/MATH] and [MATH]c_2[/MATH]

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

 

[msd1997]

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

like this

[MATH]f(x) = c_1 e^x \cos x + c_2 e^x \sin x[/MATH]
[MATH]e^x[/MATH] will take the real number and [MATH]\cos x[/MATH] and [MATH]\sin x[/MATH] will take the imaginary number

you have [MATH]m = 1 + 3i[/MATH] and [MATH]m = 1 - 3i[/MATH]
this means you have [MATH]1[/MATH] as real number and [MATH]3i[/MATH] as an imaginary number

so the solution will be

[MATH]f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x[/MATH]
Now, you can continue with the conditions to find [MATH]c_1[/MATH] and [MATH]c_2[/MATH]

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

 

[Deleted member 4993]

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]

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 [MATH]-3i[/MATH] and [MATH]3i[/MATH] as just [MATH]3i[/MATH]
Why?

because [MATH]1+3i[/MATH], will give you this equation
[MATH]y_1(x) = e^x \cos 3x + e^x \sin 3x [/MATH]
and

[MATH]1-3i[/MATH], will give you this equation
[MATH]y_2(x) = e^x \cos -3x + e^x \sin -3x = e^x \cos 3x - e^x \sin 3x[/MATH]
the solution is their combination

[MATH]u(x) = y_1(x) + y_2(x) = 2e^x \cos 3x [/MATH][MATH]u(x) = \frac{1}{2}y_1(x) + \frac{1}{2}y_2(x) = e^x \cos 3x [/MATH]
we can have another solution as
[MATH]v(x) = y_1(x) - y_2(x) = 2ie^x \sin 3x [/MATH][MATH]v(x) = \frac{1}{2i}y_1(x) - \frac{1}{2i}y_2(x) = e^x \sin 3x [/MATH]
the general solution is just a combination of the two

[MATH]y(x) = c_1 u(x) + c_2 v(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x[/MATH]

 

[msd1997]

You can always ignore the negative value and treat [MATH]-3i[/MATH] and [MATH]3i[/MATH] as just [MATH]3i[/MATH]
Why?

because [MATH]1+3i[/MATH], will give you this equation
[MATH]y_1(x) = e^x \cos 3x + e^x \sin 3x [/MATH]
and

[MATH]1-3i[/MATH], will give you this equation
[MATH]y_2(x) = e^x \cos -3x + e^x \sin -3x = e^x \cos 3x - e^x \sin 3x[/MATH]
the solution is their combination

[MATH]u(x) = y_1(x) + y_2(x) = 2e^x \cos 3x [/MATH][MATH]u(x) = \frac{1}{2}y_1(x) + \frac{1}{2}y_2(x) = e^x \cos 3x [/MATH]
we can have another solution as
[MATH]v(x) = y_1(x) - y_2(x) = 2ie^x \sin 3x [/MATH][MATH]v(x) = \frac{1}{2i}y_1(x) - \frac{1}{2i}y_2(x) = e^x \sin 3x [/MATH]
the general solution is just a combination of the two

[MATH]y(x) = c_1 u(x) + c_2 v(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x[/MATH]

Can you please tell me if my solution is correct?

 

[msd1997]

Is this correct?

 

[nasi112]

Your derivative is wrong

the derivative of

[MATH]f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x[/MATH]
Is Not

[MATH]f'(x) = c_1 e^x(-3\sin 3x) + c_2 e^x (3\cos 3x)[/MATH]
Did you forget the product rule when you have two functions multiplying by each other such as [MATH]e^x \cos 3x[/MATH]?

 

[msd1997]

Your derivative is wrong

the derivative of

[MATH]f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x[/MATH]
Is Not

[MATH]f'(x) = c_1 e^x(-3\sin 3x) + c_2 e^x (3\cos 3x)[/MATH]
Did you forget the product rule when you have two functions multiplying by each other such as [MATH]e^x \cos 3x[/MATH]?

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

 

[nasi112]

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.