- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

$$\dfrac{dy}{dx}=2\sin x$$

with the initial condition

$$y(\pi)=1$$

a. $y=2\cos{x}+3$

b. $y=2\cos{x}-1$

c. $y=-2\cos{x}+3$

d. $y=-2\cos{x}+1$

e. $y=-2\cos{x}-1$

integrate

$y=\displaystyle\int 2\sin x\, dx =-2\cos(\pi)+C$

then plug in $y(\pi)=1$

$-2\cos(\pi)+C=1

\Rightarrow

-2(-1)+C=1

\Rightarrow

C=-1$

therefore

$y=-2\cos(\pi)-1$

which is e

typos maybe!!!