MHB What is the Solution to the Differential Equation dy/dx = (2cos 2x)/(3+2y)?

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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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You now need to use the given point to get the particular solution, and then answer the rest of the question.
 
(-1)^2 + 3 (-1)= sin(2 (0)) + c
1-3=1+c
-3=c
 
$y^2+3y=\sin 2x -3$
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 don't think this is headed towards the answer!
 
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the book ans is #25

how did they get 1/4
2020_05_06_10.28.47~2.jpg
 
$\sin(2\cdot 0) = 0 \implies C = -2$
 
Rather than complete the square (which works just fine), you can also use the quadratic formula:

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

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

Using the given initial value, we find:

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

$$-3\pm\sqrt{9+4c_1}=-2$$

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

$$\sqrt{9+4c_1}=1\implies c_1=-2$$

Hence:

$$y(x)=\frac{-3+\sqrt{9+4(\sin(2x)-2)}}{2}=\frac{-3+\sqrt{4\sin(2x)+1}}{2}$$
 
well that was very helpful
mahalo
 
$$y(x)=\dfrac{-3+\sqrt{9+4(\sin(2x)-2)}}{2}
=\dfrac{-3+\sqrt{4\sin(2x)+1}}{2} =-\dfrac{3}{2}+\sqrt{\sin(2x)+\dfrac{1}{4}}$$

why is $x=\dfrac{\pi}{4}$
 
  • #10
karush said:
$$y(x)=\dfrac{-3+\sqrt{9+4(\sin(2x)-2)}}{2}
=\dfrac{-3+\sqrt{4\sin(2x)+1}}{2} =-\dfrac{3}{2}+\sqrt{\sin(2x)+\dfrac{1}{4}}$$

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

what value of x makes sin(2x) a maximum?
 
  • #11
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 [math]\frac{dy}{dx}= \frac{2 cos(2x)}{3+ 2y}[/math].

Setting that equal to 0 we immediately have 2 cos(2x)= 0. cos(a)= 0 for a any odd multiple of [math]\frac{\pi}{2}[/math] so [math]x= \frac{(2n+1)\pi}{4}[/math]. Check the actual values of y for [mATH]x= \frac{\pi}{4}[/math], [math]x= -\frac{\pi}{4}[/math], [math]x= \frac{3\pi}{4}[/math], etc. to determine the actual global maximum.
 
  • #12
skeeter said:
what value of x makes sin(2x) a maximum?
ok I see. $2(\pi/4)=\pi/2$
 

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