Why Do Solutions to This ODE Differ?

[Chipset3600]

$$\[y'sin(x)= yln(y)\]

$$
Hi, I am trying to solve this one but i can't find the same result of the book:
Here is my solution:

http://i.imgur.com/kIV2kh7.jpg

 

[Prove It]

\displaystyle \begin{align*} \frac{dy}{dx}\sin{(x)} &= y\ln{(y)} \\ \frac{1}{y\ln{(y)}}\,\frac{dy}{dx} &= \frac{1}{\sin{(x)}} \\ \frac{1}{\ln{(y)}}\,\frac{1}{y}\,\frac{dy}{dx} &= \frac{\sin{(x)}}{\sin^2{(x)}} \\ \frac{1}{\ln{(y)}}\,\frac{1}{y}\,\frac{dy}{dx} &= \frac{\sin{(x)}}{1 - \cos^2{(x)}} \\ \int{\frac{1}{\ln{(y)}}\,\frac{1}{y}\,\frac{dy}{dx}\,dx} &= \int{\frac{\sin{(x)}}{1 - \cos^2{(x)}}\,dx} \\ \int{ \frac{1}{\ln{(y)}}\,\frac{1}{y}\,dy} &= -\int{ \frac{-\sin{(x)}}{1 - \cos^2{(x)}}\,dx} \end{align*}

Now make the substitutions \displaystyle \begin{align*} u = \ln{(y)} \implies du = \frac{1}{y}\,dy \end{align*} and \displaystyle \begin{align*} v = \cos{(x)} \implies dv = -\sin{(x)}\,dx \end{align*} and the DE becomes

\displaystyle \begin{align*} \int{\frac{1}{u}\,du} &= -\int{ \frac{1}{1 - v^2}\,dv} \end{align*}

This should now be easy to solve. The RHS requires partial fractions.

 

[Chris L T521]

\displaystyle \begin{align*} \frac{dy}{dx}\sin{(x)} &= y\ln{(y)} \\ \frac{1}{y\ln{(y)}}\,\frac{dy}{dx} &= \frac{1}{\sin{(x)}} \\ \frac{1}{\ln{(y)}}\,\frac{1}{y}\,\frac{dy}{dx} &= \frac{\sin{(x)}}{\sin^2{(x)}} \\ \frac{1}{\ln{(y)}}\,\frac{1}{y}\,\frac{dy}{dx} &= \frac{\sin{(x)}}{1 - \cos^2{(x)}} \\ \int{\frac{1}{\ln{(y)}}\,\frac{1}{y}\,\frac{dy}{dx}\,dx} &= \int{\frac{\sin{(x)}}{1 - \cos^2{(x)}}\,dx} \\ \int{ \frac{1}{\ln{(y)}}\,\frac{1}{y}\,dy} &= -\int{ \frac{-\sin{(x)}}{1 - \cos^2{(x)}}\,dx} \end{align*}

Now make the substitutions \displaystyle \begin{align*} u = \ln{(y)} \implies du = \frac{1}{y}\,dy \end{align*} and \displaystyle \begin{align*} v = \cos{(x)} \implies dv = -\sin{(x)}\,dx \end{align*} and the DE becomes

\displaystyle \begin{align*} \int{\frac{1}{u}\,du} &= -\int{ \frac{1}{1 - v^2}\,dv} \end{align*}

This should now be easy to solve. The RHS requires partial fractions.

Alternatively, you could have that

$$\begin{aligned} \frac{1}{y\ln y}\frac{dy}{dx} &= \csc x\\ \frac{1}{y\ln y}\frac{dy}{dx} &= \csc x \cdot \frac{\csc x + \cot x}{\csc x+\cot x}\\ \frac{1}{y\ln y}\frac{dy}{dx} &= \frac{\csc^2x + \csc x\cot x}{\csc x + \cot x}\\ \int\frac{1}{y\ln y}\,dy &= -\int\frac{-\csc^2x-\csc x\cot x}{\csc x+\cot x}\,dx\end{aligned}$$

Making the substitutions $u=\ln y\implies \,du=\frac{1}{y}\,dy$ and $v=\csc x+\cot x \implies \,dv = -\csc^2x-\csc x\cot x\,dx$ transforms the problem into
$$\int\frac{1}{u}\,du = -\int\frac{1}{v}\,dv$$
which is an easy problem to solve from here. The reason why I also present it this way is because the OP seems to be relying on the fact that $\displaystyle\int \csc x \,dx = -\ln|\csc x + \cot x| + C$ in the solution attempt seen in their original post.

 

[Chipset3600]

Actually the problem isn't solve the integral, as you can see in my link "

http://i.imgur.com/kIV2kh7.jpg

" with my solution i found a particular solution that isn't the same of book: "y = 1".