Solve-dy-dx-2cos-2x-3-2y-with-y-0-1-i-for-what-value-of-x-gt-0-does-the-situation-exist-ii-for-what-value-of-x-is-y-x-maximum-

Question Number 17323 by tawa tawa last updated on 04/Jul/17

Solve : \frac{dy}{dx} = \frac{2 \cos (2 x)}{3 - 2 y} with y (0) = - 1

(i) for what value of x > 0 does the situation exist

(ii) for what value of x is y (x) maximum

Answered by Tinkutara last updated on 04/Jul/17

(3 - 2 y) d y = 2 \cos 2 x d x

\int (3 - 2 y) d y = 2 \int \cos 2 x d x

3 y - y^{2} = \sin 2 x + C

- 3 - 1 = C = - 4

∴ 3 y - y^{2} = \sin 2 x - 4

Commented by tawa tawa last updated on 04/Jul/17

God bless you sir . is that all ?

Answered by mrW1 last updated on 04/Jul/17

(3 - 2 y) dy = 2 \cos (2 x) dx

\int (3 - 2 y) dy = 2 \int \cos (2 x) dx

- \frac{1}{2} \int (3 - 2 y) d (3 - 2 y) = \int \cos (2 x) d (2 x)

- \frac{1}{4} {(3 - 2 y)}^{2} = \sin (2 x) + C

y (0) = - 1

- \frac{1}{4} {(3 + 2)}^{2} = C

C = - \frac{25}{4}

\frac{25}{4} - \frac{1}{4} {(3 - 2 y)}^{2} = \sin (2 x)

[5^{2} - {(3 - 2 y)}^{2}] = \sin (2 x)

(5 - 3 + 2 y) (5 + 3 - 2 y) = \sin (2 x)

4 (1 + y) (4 - y) = \sin (2 x)

4 (\frac{25}{4} + - \frac{9}{4} + 3 y - y^{2}) = \sin (2 x)

25 - 4 {(y - \frac{3}{2})}^{2} = \sin (2 x)

{(y - \frac{3}{2})}^{2} = \frac{25 - \sin (2 x)}{4}

y = \frac{1}{2} [3 \pm 5 \sqrt{1 - \frac{\sin (2 x)}{25}}]

(a) for x \in R

(b) y = maximum / minimum when \sin (2 x) = - 1

or 2 x = 2 k π + \frac{3 π}{2}

or x = k π + \frac{3 π}{4}

Commented by tawa tawa last updated on 04/Jul/17

Wow God bless you sir .

Commented by 1234Hello last updated on 04/Jul/17

Sir, but why in 3^{rd} line you integrate

with respect to the function and not

x ? Sir simply \int (3 - 2 y) d y = 3 y - y^{2} ?

Commented by Tinkutara last updated on 04/Jul/17

Then my answer is correct or not ?

Commented by mrW1 last updated on 04/Jul/17

there is no special reason . it makes no

real difference . i just wanted to keep

the terms together .

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