Question

I have solved for the differential equation, I just don't know how to get the final answer..

-ycosy+siny = -sin(x^-1) +c initial condition: y(1/pi) = pi

What do I plug in where and what is my final answer in exact form?

Answer 1

Think of it this way: If you had the solution \(y=x+C\) (for simplicity) with the initial value \(y(0)=1\), you would then replace every \(y\) you see with \(1\) and every \(x\) with 0.

The same can be done for your equation. If \(y\left(\dfrac{1}{\pi}\right)=\pi\), then
\[-\pi\cos\pi+\sin\pi=-\sin\left(\frac{1}{\pi}\right)^{-1}+C\]

Answer 2

what is -sin(1/pi) inverse equal to?

Answer 3

I need an exact form

Answer 4

so far I have -pi= sin(1/Pi) +c

Answer 5

Well, firstly you have
\[-\sin\bigg[\left(\frac{1}{\pi}\right)^{-1}\bigg]\]
which means you have the reciprocal of \(\dfrac{1}{\pi}\), or just \(\pi\):
\[-\sin\pi\]
Now what's the sine of \(\pi\)? What's the negative of that?

Answer 6

wait a minute something is up

Answer 7

shouldn't it be sin (pi^-1)

Answer 8

that becomes sin (1/pi)

Answer 9

No, the initial condition is \(y\left(\dfrac{1}{\pi}\right)=\pi\), meaning \(x=\dfrac{1}{\pi}\) when \(y=\pi\).

The \(x^{-1}\) in the general solution is equivalent to \(\dfrac{1}{x}\), so
\[\frac{1}{1/\pi}=\pi\]

Answer 10

so what would be my final answer for C, I just want to be sure I got it right.

Answer 11

\[-\pi\cos\pi+\sin\pi=-\sin\left(\frac{1}{\pi}\right)^{-1}+C\]
reduces to
\[-\pi(-1)+0=-\sin\pi+C\]
then
\[\pi=0+C~~\iff~~C=\pi\]

Answer 12

oh I got -1 but I C how you got C.

Answer 13

Thank you best responce

Answer 14

You're welcome!