MHB Apc.9.3.1 solution to the differential equation condition

[karush]

253 Which of the following is the solution to the differential equation condition
$$\dfrac{dy}{dx}=2\sin x$$
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 etypos maybe!

 

[skeeter]

253 Which of the following is the solution to the differential equation condition
$$\dfrac{dy}{dx}=2\sin x$$
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
$y=\displaystyle\int 2\sin x\, dx =\color{red}-2\cos{x}+C$
then plug in $y(\pi)=1$
$-2\cos(\pi)+C=1
\Rightarrow
-2(-1)+C=1
\Rightarrow
C=-1$
therefore
$\color{red}y=-2\cos{x}-1$
which is etypos maybe!

​yep

 

[karush]