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anonymous
 5 years ago
How about this one?,,,y'sinx=yIny
anonymous
 5 years ago
How about this one?,,,y'sinx=yIny

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This one is a bit more involved. It is again separable, so\[\frac{dy}{y.\ln(y)}=\frac{dx}{\sin(x)}\] The lefthand side can be integrated using the substitution u=ln(y), for then du=(1/y)dy and we can rewrite the lefthand side as\[\frac{du}{u} \rightarrow \int\limits_{}{}\frac{du}{u}=\ln(u)+c_1\]Substituting u=ln(y) back in, we get\[\ln(\ln(y))+c_1\]as solution to the lefthand side. I'll post the righthand side next...I have to break it up since this site is awkward to use when typing large pieces.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to use a tsubstitution on the righthand side. I was going to go into setting it up, but this site is difficult to use. I'm going to assume you have heard of the tsubstitution (if not, it's easy to find out information). From this substitution technique, we have two results to use:\[\sin(x)=\frac{2t}{1+t^2}\] and \[dx=\frac{2dt}{1+t^2}\]Plugging these into dx/sin(x) we have\[\int_{}{}\frac{dx}{\sin(x)} \rightarrow \int\limits_{}{}\frac{1+t^2}{2t}.\frac{2dt}{1+t^2}=\int\limits_{}{}\frac{dt}{t}=\ln(t)+c_2\]Now, from the definition of tsubstitution, \[t=\tan(\frac{x}{2})\]so upon substituting, the righthand side becomes\[\ln(\tan(\frac{x}{2}))+c_2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0am enjoying your leasons ,sir

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, now we bring the two answers together,\[\ln(\ln(y))+c_1=\ln(\tan(\frac{x}{2}))+c_2\]Collecting the constants on the RHS and exponentiating, we get,\[e^{\ln(\ln(y))}=e^{\ln(\tan(\frac{x}{2}))+(c_2c_1)}\]Which simplifies to\[\ln(y)=e^{(c_2c_1)}e^{\ln(\tan(\frac{x}{2}))}=A{\tan(\frac{x}{2})}\]where A is just e^(c_2c_1), a constant. Exponentiating again isolates y,\[e^{\ln(y)}=e^{Atan(\frac{x}{2})}=e^Ae^{\tan(\frac{x}{2})}=Be^{\tan(\frac{x}{2})}\]where again, B is just the collected constant, e^A. So you're answer is \[y=Be^{\tan(\frac{x}{2})}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Grammatical error  *your* answer, not, *you're*.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Remember to check everything. You can do that by differentiating the final answer  you should be able to get back to the original equation. I haven't checked anything I've sent, so please do that. Good luck. Bed time...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0PS  I'm glad you enjoyed the responses!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow!!))thanx very much, sir,, i have really enjoyed your responses ,
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