# Do not know how to handle complex numbers

#### msd1997

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

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#### nasi112

##### Full Member
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

##### New member
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

##### New member
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
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

##### Super Moderator
Staff member
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

##### Full Member
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

##### New member
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?

Is this correct?

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#### nasi112

##### Full Member

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

##### New member

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

##### Full Member
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.