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DE 2.2.25 IVP - Find solution and determine where maximum occurs

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
2020_05_05_22.40.35~2.jpg
it's late so I'll just start this
\[ \dfrac{dy}{dx}=\dfrac{2\cos 2x}{3+2y} \]
so \[(3+2y) \, dy= (2\cos 2x) \, dx\]
$y^2 + 3 y= sin(2 x) + c$
 
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Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
You now need to use the given point to get the particular solution, and then answer the rest of the question.
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
(-1)^2 + 3 (-1)= sin(2 (0)) + c
1-3=1+c
-3=c
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
$y^2+3y=\sin 2x -3$
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!
 
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karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
the book ans is #25

how did they get 1/42020_05_06_10.28.47~2.jpg
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
$\sin(2\cdot 0) = 0 \implies C = -2$
 

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,661
St. Augustine, FL.
Rather than complete the square (which works just fine), you can also use the quadratic formula:

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

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

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:

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

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
well that was very helpful
mahalo
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
\(\displaystyle y(x)=\dfrac{-3+\sqrt{9+4(\sin(2x)-2)}}{2}
=\dfrac{-3+\sqrt{4\sin(2x)+1}}{2} =-\dfrac{3}{2}+\sqrt{\sin(2x)+\dfrac{1}{4}}\)

why is $x=\dfrac{\pi}{4}$
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
\(\displaystyle y(x)=\dfrac{-3+\sqrt{9+4(\sin(2x)-2)}}{2}
=\dfrac{-3+\sqrt{4\sin(2x)+1}}{2} =-\dfrac{3}{2}+\sqrt{\sin(2x)+\dfrac{1}{4}}\)

why is $x=\dfrac{\pi}{4}$
what value of x makes sin(2x) a maximum?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Although I am normally an advocate of "completing the square" to find max or min, here I advise setting the derivative of y equal to 0. That becomes especially easy because you are given that \(\displaystyle \frac{dy}{dx}= \frac{2 cos(2x)}{3+ 2y}\).

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.
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii