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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-

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Question Number 17323 by tawa tawa last updated on 04/Jul/17
Solve: (dy/dx) = ((2cos(2x))/(3 − 2y)) 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
$$\mathrm{Solve}:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{2cos}\left(\mathrm{2x}\right)}{\mathrm{3}\:−\:\mathrm{2y}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{with}\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)\:=\:−\mathrm{1} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{for}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:>\:\mathrm{0}\:\mathrm{does}\:\mathrm{the}\:\mathrm{situation}\:\mathrm{exist} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{for}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{is}\:\mathrm{y}\left(\mathrm{x}\right)\:\mathrm{maximum} \\ $$
Answered by Tinkutara last updated on 04/Jul/17
(3 − 2y) dy = 2 cos 2x dx ∫(3 − 2y) dy = 2∫cos 2x dx 3y − y^2 = sin 2x + C −3 − 1 = C = −4 ∴ 3y − y^2 = sin 2x − 4
$$\left(\mathrm{3}\:−\:\mathrm{2}{y}\right)\:{dy}\:=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{x}\:{dx} \\ $$$$\int\left(\mathrm{3}\:−\:\mathrm{2}{y}\right)\:{dy}\:=\:\mathrm{2}\int\mathrm{cos}\:\mathrm{2}{x}\:{dx} \\ $$$$\mathrm{3}{y}\:−\:{y}^{\mathrm{2}} \:=\:\mathrm{sin}\:\mathrm{2}{x}\:+\:{C} \\ $$$$−\mathrm{3}\:−\:\mathrm{1}\:=\:{C}\:=\:−\mathrm{4} \\ $$$$\therefore\:\mathrm{3}{y}\:−\:{y}^{\mathrm{2}} \:=\:\mathrm{sin}\:\mathrm{2}{x}\:−\:\mathrm{4} \\ $$
Commented by tawa tawa last updated on 04/Jul/17
God bless you sir. is that all ?
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{is}\:\mathrm{that}\:\mathrm{all}\:? \\ $$
Answered by mrW1 last updated on 04/Jul/17
(3−2y)dy=2cos (2x)dx ∫(3−2y)dy=2∫cos (2x)dx −(1/2)∫(3−2y)d(3−2y)=∫cos (2x)d(2x) −(1/4)(3−2y)^2 =sin (2x)+C y(0)=−1 −(1/4)(3+2)^2 =C C=−((25)/4) ((25)/4)−(1/4)(3−2y)^2 =sin (2x) [5^2 −(3−2y)^2 ]=sin (2x) (5−3+2y)(5+3−2y)=sin (2x) 4(1+y)(4−y)=sin (2x) 4(((25)/4)+−(9/4)+3y−y^2 )=sin (2x) 25−4(y−(3/2))^2 =sin (2x) (y−(3/2))^2 =((25−sin (2x))/4) y=(1/2)[3±5(√(1−((sin (2x))/(25))))] (a)for x∈R (b)y=maximum/minimum when sin (2x)=−1 or 2x=2kπ+((3π)/2) or x=kπ+((3π)/4)
$$\left(\mathrm{3}−\mathrm{2y}\right)\mathrm{dy}=\mathrm{2cos}\:\left(\mathrm{2x}\right)\mathrm{dx} \\ $$$$\int\left(\mathrm{3}−\mathrm{2y}\right)\mathrm{dy}=\mathrm{2}\int\mathrm{cos}\:\left(\mathrm{2x}\right)\mathrm{dx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{3}−\mathrm{2y}\right)\mathrm{d}\left(\mathrm{3}−\mathrm{2y}\right)=\int\mathrm{cos}\:\left(\mathrm{2x}\right)\mathrm{d}\left(\mathrm{2x}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}−\mathrm{2y}\right)^{\mathrm{2}} =\mathrm{sin}\:\left(\mathrm{2x}\right)+\mathrm{C} \\ $$$$\mathrm{y}\left(\mathrm{0}\right)=−\mathrm{1} \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{C} \\ $$$$\mathrm{C}=−\frac{\mathrm{25}}{\mathrm{4}} \\ $$$$\frac{\mathrm{25}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}−\mathrm{2y}\right)^{\mathrm{2}} =\mathrm{sin}\:\left(\mathrm{2x}\right) \\ $$$$\left[\mathrm{5}^{\mathrm{2}} −\left(\mathrm{3}−\mathrm{2y}\right)^{\mathrm{2}} \right]=\mathrm{sin}\:\left(\mathrm{2x}\right) \\ $$$$\left(\mathrm{5}−\mathrm{3}+\mathrm{2y}\right)\left(\mathrm{5}+\mathrm{3}−\mathrm{2y}\right)=\mathrm{sin}\:\left(\mathrm{2x}\right) \\ $$$$\mathrm{4}\left(\mathrm{1}+\mathrm{y}\right)\left(\mathrm{4}−\mathrm{y}\right)=\mathrm{sin}\:\left(\mathrm{2x}\right) \\ $$$$\mathrm{4}\left(\frac{\mathrm{25}}{\mathrm{4}}+−\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{3y}−\mathrm{y}^{\mathrm{2}} \right)=\mathrm{sin}\:\left(\mathrm{2x}\right) \\ $$$$\mathrm{25}−\mathrm{4}\left(\mathrm{y}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{sin}\:\left(\mathrm{2x}\right) \\ $$$$\left(\mathrm{y}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{25}−\mathrm{sin}\:\left(\mathrm{2x}\right)}{\mathrm{4}} \\ $$$$\mathrm{y}=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{3}\pm\mathrm{5}\sqrt{\mathrm{1}−\frac{\mathrm{sin}\:\left(\mathrm{2x}\right)}{\mathrm{25}}}\right] \\ $$$$ \\ $$$$\left(\mathrm{a}\right)\mathrm{for}\:\mathrm{x}\in\mathrm{R} \\ $$$$\left(\mathrm{b}\right)\mathrm{y}=\mathrm{maximum}/\mathrm{minimum}\:\mathrm{when}\:\mathrm{sin}\:\left(\mathrm{2x}\right)=−\mathrm{1} \\ $$$$\mathrm{or}\:\mathrm{2x}=\mathrm{2k}\pi+\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$\mathrm{or}\:\mathrm{x}=\mathrm{k}\pi+\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$
Commented by tawa tawa last updated on 04/Jul/17
Wow God bless you sir.
$$\mathrm{Wow}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$
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 ∫(3 − 2y) dy = 3y − y^2 ?
$$\mathrm{Sir},\:\mathrm{but}\:\mathrm{why}\:\mathrm{in}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{line}\:\mathrm{you}\:\mathrm{integrate} \\ $$$$\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{function}\:\mathrm{and}\:\mathrm{not} \\ $$$${x}?\:\mathrm{Sir}\:\mathrm{simply}\:\int\left(\mathrm{3}\:−\:\mathrm{2}{y}\right)\:{dy}\:=\:\mathrm{3}{y}\:−\:{y}^{\mathrm{2}} ? \\ $$
Commented by Tinkutara last updated on 04/Jul/17
Then my answer is correct or not?
$$\mathrm{Then}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{or}\:\mathrm{not}? \\ $$
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.
$$\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{special}\:\mathrm{reason}.\:\mathrm{it}\:\mathrm{makes}\:\mathrm{no} \\ $$$$\mathrm{real}\:\mathrm{difference}.\:\mathrm{i}\:\mathrm{just}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{keep} \\ $$$$\mathrm{the}\:\mathrm{terms}\:\mathrm{together}. \\ $$

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