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Question Number 43008 by MJS last updated on 06/Sep/18

(y′)^2 =−1+sin x y=?

$$\left({y}'\right)^{\mathrm{2}} =−\mathrm{1}+\mathrm{sin}\:{x} \\ $$$${y}=? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18

(dy/dx)=(√(−1+sinx)) (dy/dx)=(√(−1(1−sinx))) (dy/dx)=i(√(sin^2 (x/2)−2sin(x/2)cos(x/2)+cos^2 (x/2))) (dy/dx)=i(sin(x/2)−cos(x/2)) dy=i(sin(x/2)−cos(x/2))dx y=i(−2cos(x/2)−2sin(x/2))+c← ans recheck y=−2i(cos(x/2)+sin(x/2)) (dy/dx)=−2i(−(1/2)sin(x/2)+(1/2)cos(x/2)) (dy/dx)=i(sin(x/2)−cos(x/2)) ((dy/dx))^2 =−1(1−sinx) ((dy/dx))^2 =(−1+sinx) is it right or wrong pls check...

$$\frac{{dy}}{{dx}}=\sqrt{−\mathrm{1}+{sinx}}\: \\ $$$$\frac{{dy}}{{dx}}=\sqrt{−\mathrm{1}\left(\mathrm{1}−{sinx}\right)}\: \\ $$$$\frac{{dy}}{{dx}}={i}\sqrt{{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}+{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$$$\frac{{dy}}{{dx}}={i}\left({sin}\frac{{x}}{\mathrm{2}}−{cos}\frac{{x}}{\mathrm{2}}\right)\:\: \\ $$$${dy}={i}\left({sin}\frac{{x}}{\mathrm{2}}−{cos}\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$${y}={i}\left(−\mathrm{2}{cos}\frac{{x}}{\mathrm{2}}−\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}\right)+{c}\leftarrow\:{ans} \\ $$$${recheck} \\ $$$${y}=−\mathrm{2}{i}\left({cos}\frac{{x}}{\mathrm{2}}+{sin}\frac{{x}}{\mathrm{2}}\right) \\ $$$$\frac{{dy}}{{dx}}=−\mathrm{2}{i}\left(−\frac{\mathrm{1}}{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}\right) \\ $$$$\frac{{dy}}{{dx}}={i}\left({sin}\frac{{x}}{\mathrm{2}}−{cos}\frac{{x}}{\mathrm{2}}\right) \\ $$$$\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =−\mathrm{1}\left(\mathrm{1}−{sinx}\right) \\ $$$$\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\left(−\mathrm{1}+{sinx}\right) \\ $$$${is}\:{it}\:{right}\:{or}\:{wrong}\:{pls}\:{check}... \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by abdo.msup.com last updated on 06/Sep/18

its clear that y is a complex function you answer is correct you have verifying this.

$${its}\:{clear}\:{that}\:\:{y}\:{is}\:{a}\:{complex}\:{function} \\ $$$${you}\:{answer}\:{is}\:{correct}\:{you}\:{have}\:{verifying} \\ $$$${this}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18

thank you sir...

$${thank}\:{you}\:{sir}... \\ $$

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