Beginning calculus trig-related integral

So it starts like this:

**Find the derivative of y = isinx + cosx**

And I don't know what I'm supposed to do with the complex number so I just sort of left it and got this:

dy/dx = -sinx + icosx

Which is equal to i * y

Then it says to

1) find dy/y

2) use the condition x=0 to determine the value of *c* (constant of integration)

3) write this antiderivative in its equivalent form

I don't know what to do next! Please dumb it down as much as possible. :D

Re: Beginning calculus trig-related integral

Since you have $\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} = i\,y \end{align*}$, what does that mean $\displaystyle \displaystyle \begin{align*} \frac{dy}{y} \end{align*}$ is?

Re: Beginning calculus trig-related integral

Re: Beginning calculus trig-related integral

Yes, so now that you have $\displaystyle \displaystyle \begin{align*} \frac{dy}{y} = i\,dx \end{align*}$ you should be able to integrate both sides.

Re: Beginning calculus trig-related integral

Re: Beginning calculus trig-related integral

First of all, $\displaystyle \displaystyle \begin{align*} \int{\frac{dy}{y}} = \ln{|y|} \end{align*}$, not $\displaystyle \displaystyle \begin{align*} \ln{(y)} \end{align*}$. And what would be $\displaystyle \displaystyle \begin{align*} \int{i\,dx} \end{align*}$? Remember that $\displaystyle \displaystyle \begin{align*} i \end{align*}$ is a constant.

Re: Beginning calculus trig-related integral

Re: Beginning calculus trig-related integral

So boundary condition x=0

y = isin(0) + cos(0)

y = 1

ln|1| = i(0) + c

c = 0

Is this right? What does it mean to write the antiderivative in its equivalent form?

Re: Beginning calculus trig-related integral

So if

ln|y| = ix + 0

then would the equivalent form be

y = e^(ix)

Re: Beginning calculus trig-related integral

Quote:

Originally Posted by

**euphony** So if

ln|y| = ix + 0

then would the equivalent form be

y = e^(ix)

That is the correct equivalent form, however, your elimination of the constant is incorrect due to the absolute value around the y.

$\displaystyle \displaystyle \begin{align*} \ln{|y|} &= i\,x + C \\ |y| &= e^{i\,x + C} \\ |y| &= e^C e^{i\,x} \\ y &= A\,e^{i\,x} \end{align*}$

Now use your boundary condition to evaluate this constant A.

Re: Beginning calculus trig-related integral

y = Ae^(ix)

(1) = A * e^[i(0)]

A = 1?

Re: Beginning calculus trig-related integral