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[SOLVED] 9.3.253 AP Calculus Exam .... solution to the differential equation condition

karush

Well-known member
Jan 31, 2012
2,724
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 e


typos maybe!!!
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
702
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 e


typos maybe!!!
​yep
 

karush

Well-known member
Jan 31, 2012
2,724
:cool: