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Solve the following differential question:

$\displaystyle \sin^{-1}\left(\frac{dy}{dx}\right )=x+y$

My Attempt:

$\displaystyle \frac{dy}{dx}=\sin(x+y)$ $$\begin{align} \displaystyle Put\; x+y=t \\ \displaystyle 1+\frac{dy}{dx}=\frac{dt}{dx} \\ \displaystyle \frac{dt}{dx}-1=\sin t \\ \displaystyle \frac{dt}{dx}=1+\sin t \\ \displaystyle \frac{dt}{1+\sin t}=dx \\ \end{align}$$ Integrating

$$\begin{align} \displaystyle \int \frac{dt}{1+\sin t}=\int dx \end{align}$$ Divide both numerator and denominator by $\displaystyle \cos t $ $$\begin{align} \displaystyle \int \frac{\sec t dt}{\sec t+\tan t}=\int dx \end{align}$$ Put $\displaystyle \sec t+\tan t=z $ $$\begin{align} \displaystyle \sec t\tan t dt+\sec^2 t dt=dz \\ \displaystyle \sec t(\tan t+\sec t) dt=dz \\ \displaystyle \sec t dt=\frac{dz}{z} \\ \end{align}$$

So the integral becomes $\displaystyle \int \frac{dz}{z^2}=\int dx $

$$\begin{align} \displaystyle -\frac{1}{z}=x+c \\ \displaystyle -\frac{1}{\tan t+\sec t}=x+c \\ \displaystyle -\frac{1}{\tan(x+y)+\sec(x+y)}=x+c \\ \end{align}$$

I am stuck here. The given solution is $\displaystyle tan(x+y)=sec(x+y)+x+c $

Please help.

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  • $\begingroup$ You are close. Use $\int\frac{dt}{1+\sin t}=\frac2{\cot\frac t2+1}$, which will allow you to find an explicit form $y=f(x)$. $\endgroup$ – Yves Daoust Jun 4 '14 at 12:22
  • $\begingroup$ I dint get this .. $\endgroup$ – square_one Jun 4 '14 at 12:49
  • $\begingroup$ The given solution is an implicit equation $f(x,y)=0$, not so handy. $\endgroup$ – Yves Daoust Jun 4 '14 at 13:04
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You did the hardest and I'm almost ashamed to write an answer: \begin{eqnarray*} -\frac{1}{\tan(t) + \sec(t)} &=& -\frac{\sec(t) - \tan(t)}{\sec^2(t) - \tan^2(t)}\\ &=& \cos^2(t)\frac{\tan(t) - \sec(t)}{1 - \sin^2(t)}\\ &=& \tan(t) - \sec(t) \end{eqnarray*}

Letting $t = x+y$ yields your answer.

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