I have a differential equation:

$$ \frac{dy}{dx} = y \log(y)\cot(x)$$

I'm trying solve that equation by separating variables and dividing by $y\log(y)$:

$$ dy = y \log(y) \cot(x) dx$$

$$ \frac{dy}{y \log(y)} = \cot(x) dx$$

$$ \cot(x) - \frac{dy}{y \log(y)} = 0 $$

Where of course $ y > 0 $ regarding to division


$$ \int \frac{dy}{y \log(y)} = \ln | \ln(y) | +C $$


$$ \int \cot(x) dx = \ln| \sin(x) | + C$$


$$ \ln| \sin(x) | - \ln | \ln(y) | = C $$

$$ \ln \lvert\frac{\sin(x)} {\ln(y)} \rvert = C $$

$$ \frac{\sin(x)}{\ln(y)} = \pm e^{C} $$

$d = \pm e^{C} $

$$ \sin(x) = d \ln(y) $$

$$ \frac{\sin(x)}{d} = \ln(y)$$

$$ e^{\frac{\sin(x)}{d}} = y$$

This is my final answer. I have problem because in book from equation comes the answer to exercise is:

$$ y = e^{c \sin(x)}$$

Which one is correct?

I will be grateful for explaining Best regards

  • 3
    $\begingroup$ They're the same answer. It depends which side you put the arbitrary constant on. $\endgroup$ – Paul Apr 18 '17 at 12:00
  • 5
    $\begingroup$ The fraction $1/d$ is a constant. So is $c$. However, it appears you lost a solution since $c=0$ yields a valid solution. Thus, your answer is incomplete. The answer in the book is correct. $\endgroup$ – quasi Apr 18 '17 at 12:02
  • $\begingroup$ @Krzystof Your book's ans s correct. You must notice hat $c$ and $\ln c$ both are constants . Thus try using $\ln c$ as constant of integration. You will reach where the book specifies. $\endgroup$ – The Dead Legend Apr 18 '17 at 12:08

Note that $$ \int \frac{du}{u} = \ln |u| + c_1 = \ln |u| + \ln |c| = \ln |cu|.$$

Starting from your result:

$$ \cot x\, dx =\frac{dy}{y \ln y} $$ $$\implies \frac{d(\sin x)}{\sin x} = \frac{d(\ln y)}{y \ln y}$$

we integrate both sides:

$$ \int \frac{d(\sin x)}{\sin x} \, dx = \int \frac{d(\ln y)}{\ln y}\, dx$$ $$\implies \ln |a\sin x| = \ln |b\ln y| $$ $$\implies a\sin x = b\ln y, \quad(1)$$ $$\implies \ln y = c\sin x$$ $$\implies y = e^{c\sin x}$$ where $a, b, c$ are arbitrary constants. Note that $(1)$ includes all possible solutions including the case $a=0$.


NOTE: $$\ln|\frac{\sin x}{\ln y}|=C$$ Can also be written as $$\ln|\frac{\sin x}{\ln y}|=\ln C$$ (Because both terms are constant.)

This shall give you $$\sin x =C \ln y$$ Which will give you: $$y=e^{c \sin x}$$ where $c=\frac{1}{C}$. I will prefer this solution because in your answer d=0 won't be giving you any solution but Note that c=0 will do. {Credits to that comment}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.