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

[karush]

it's late so I'll just start this
\[ \dfrac{dy}{dx}=\dfrac{2\cos 2x}{3+2y} \]
so \[(3+2y) \, dy= (2\cos 2x) \, dx\]
$y^2 + 3 y= sin(2 x) + c$

 

[Prove It]

You now need to use the given point to get the particular solution, and then answer the rest of the question.

 

[karush]

(-1)^2 + 3 (-1)= sin(2 (0)) + c 1-3=1+c -3=c

 

[karush]

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

 

[karush]

the book ans is #25

how did they get 1/4

 

[skeeter]

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

 

[MarkFL]

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:

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

 

[karush]

well that was very helpful
mahalo

 

[karush]

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

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

 

[skeeter]

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

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

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

 

[HallsofIvy]

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.

 

[karush]

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

ok I see. $2(\pi/4)=\pi/2$