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- Thread starter karush
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(-1)^2 + 3 (-1)= sin(2 (0)) + c 1-3=1+c -3=c |

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- #4

$y^2+3y=\sin 2x -3$

complete square

$y^2+3y+\dfrac{9}{4}=\sin 2x -3+\dfrac{9}{4}=\sin 2x -\dfrac{3}{4}$

$\left(y+\dfrac{3}{2}\right)^2=\sin 2x -\dfrac{3}{4}$

$y=-\dfrac{3}{2}+\sqrt{\sin 2x -\dfrac{3}{4}}$

i dont think this is headed towards the answer!

complete square

$y^2+3y+\dfrac{9}{4}=\sin 2x -3+\dfrac{9}{4}=\sin 2x -\dfrac{3}{4}$

$\left(y+\dfrac{3}{2}\right)^2=\sin 2x -\dfrac{3}{4}$

$y=-\dfrac{3}{2}+\sqrt{\sin 2x -\dfrac{3}{4}}$

i dont think this is headed towards the answer!

Last edited:

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- #5

$\sin(2\cdot 0) = 0 \implies C = -2$

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- #7

\(\displaystyle y^2+3y-\sin(2x)-c_1=0\)

\(\displaystyle y(x)=\frac{-3\pm\sqrt{9+4(\sin(2x)+c_1)}}{2}\)

Using the given initial value, we find:

\(\displaystyle y(0)=\frac{-3\pm\sqrt{9+4c_1}}{2}=-1\)

\(\displaystyle -3\pm\sqrt{9+4c_1}=-2\)

We find we should take the positive root for \(y\):

\(\displaystyle \sqrt{9+4c_1}=1\implies c_1=-2\)

Hence:

\(\displaystyle y(x)=\frac{-3+\sqrt{9+4(\sin(2x)-2)}}{2}=\frac{-3+\sqrt{4\sin(2x)+1}}{2}\)

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- #8

well that was very helpful

mahalo

mahalo

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- #9

=\dfrac{-3+\sqrt{4\sin(2x)+1}}{2} =-\dfrac{3}{2}+\sqrt{\sin(2x)+\dfrac{1}{4}}\)

why is $x=\dfrac{\pi}{4}$

what value of x makes sin(2x) a maximum?

=\dfrac{-3+\sqrt{4\sin(2x)+1}}{2} =-\dfrac{3}{2}+\sqrt{\sin(2x)+\dfrac{1}{4}}\)

why is $x=\dfrac{\pi}{4}$

- Jan 29, 2012

- 1,151

Setting that equal to 0 we immediately have 2 cos(2x)= 0. cos(a)= 0 for a any odd multiple of \(\displaystyle \frac{\pi}{2}\) so \(\displaystyle x= \frac{(2n+1)\pi}{4}\). Check the actual values of y for \(\displaystyle x= \frac{\pi}{4}\), \(\displaystyle x= -\frac{\pi}{4}\), \(\displaystyle x= \frac{3\pi}{4}\), etc. to determine the actual global maximum.

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- #12

ok I see. $2(\pi/4)=\pi/2$what value of x makes sin(2x) a maximum?