Why Do Solutions to This ODE Differ?

[Chipset3600]

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

\)
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".

 

[blue_raver22]

Based on my calculation, the solution to this ODE PVI is y(x) = e^(sin(x)). However, I noticed that the book's solution is different (y(x) = e^(cos(x))). I double checked my work and could not find any errors. Can you please explain why our solutions differ?

Hi there,

Thank you for sharing your solution. After reviewing your work, I believe your solution is correct. The book's solution may be a typo or an alternate form of the solution. It is common to have multiple forms for the same solution in mathematics. As long as your solution satisfies the original ODE, it is considered correct.

Great job on solving this ODE PVI! Keep up the good work.