Question

QUIZ:

Answer 1

Solve:
\[\frac{ dy }{ dx } = \frac{ 2\cos 2x }{ 3+2y }\]
with y(0) =-1
(i) For what values of 𝑥 > 0 does the solution exist?
(ii) For what value of 𝑥 > 0 is 𝑦(𝑥) maximum?

Answer 2

@PaxPolaris any idea??

Answer 3

fill in the blanks in you questions ...

Answer 4

\[(3+2y)dy=(2\cos(2x))dx\]

Answer 5

(i) For what values of 𝑥___ > 0 does the solution exist?
(ii) For what value of 𝑥___ > 0 is 𝑦(___) maximum?

can i assume it's x, x and y?

Answer 6

yes of course., please

Answer 7

integrate both sides:\[\implies \int\limits (3+2y)dy=\int\limits 2\cos(2x)dx\]

Answer 8

𝑦2 + 3𝑦 + 2 = sin 2𝑥

right??? then??

Answer 9

\[\large y^2+3y+2=\sin(2x)\]

?

Answer 10

because y(0) = -1, so

\[\int\limits_{-1}^{y} 3 + 2y dy = \int\limits_{0}^{x} 2 \cos 2x dx\]

Answer 11

solve it
3+2y(-2xin2x.co2x) - (2cos2x)2 divided by (3+2y)^2
u ll get the ans

Answer 12

the above equation came by differentiating the equation

Answer 13

\[\Large \implies y= {-3 \pm \sqrt{9-4(2-\sin(2x)}\over 2}={-3 \pm \sqrt{1+4\sin(2x)}\over 2}\]

Answer 14

(i) y has solution when:\[\Large 1+4\sin(2x) \ge 0\]

Answer 15

ps. i just used the quadratic formula

Answer 16

then??

Answer 17

you could just use wolfram alpha for the answer