# DE 2.2.25 IVP - Find solution and determine where maximum occurs

#### karush

##### Well-known member

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

#### karush

##### Well-known member
$y^2+3y=\sin 2x -3$
complete square
$y^2+3y+\dfrac{9}{4}=\sin 2x -3+\dfrac{9}{4}=\sin 2x -\dfrac{3}{4}$
$\left(y+\dfrac{3}{2}\right)^2=\sin 2x -\dfrac{3}{4}$
$y=-\dfrac{3}{2}+\sqrt{\sin 2x -\dfrac{3}{4}}$

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#### karush

##### Well-known member
the book ans is #25

how did they get 1/4

#### skeeter

##### Well-known member
MHB Math Helper
$\sin(2\cdot 0) = 0 \implies C = -2$

#### MarkFL

##### Pessimist Singularitarian
Staff member
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:

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

mahalo

#### karush

##### Well-known member
$$\displaystyle y(x)=\dfrac{-3+\sqrt{9+4(\sin(2x)-2)}}{2} =\dfrac{-3+\sqrt{4\sin(2x)+1}}{2} =-\dfrac{3}{2}+\sqrt{\sin(2x)+\dfrac{1}{4}}$$

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

#### skeeter

##### Well-known member
MHB Math Helper
$$\displaystyle y(x)=\dfrac{-3+\sqrt{9+4(\sin(2x)-2)}}{2} =\dfrac{-3+\sqrt{4\sin(2x)+1}}{2} =-\dfrac{3}{2}+\sqrt{\sin(2x)+\dfrac{1}{4}}$$

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

#### HallsofIvy

##### Well-known member
MHB Math Helper
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
what value of x makes sin(2x) a maximum?
ok I see. $2(\pi/4)=\pi/2$