2
$\begingroup$

Actually, it is supposed to be First Order Ordinary Linear DE. I was not able to make it really linear with all my transformations. I was given a clue that I need to make substitution $z=\sin{y}$, I think it should lead to linear equation relative to $\sin{y}$. But I stuck at the very begining of substitution, how should I make it for $y$, to dissappear?

$\endgroup$
6
$\begingroup$

$$ y'\cos y + \sin y = x $$ note that $$ \dfrac{d}{dx}\sin y = y' \cos y $$ use a change of variables.

$\endgroup$
  • $\begingroup$ Thx. It is really something I should have noticed. $\endgroup$ – blitzar787 Aug 5 '15 at 16:06
  • $\begingroup$ It is something you see after years of being taught and then teaching. Though, given the hint (which i did not see on first glance of post) you ought to play around with the derivatives by inserting in the $z = \sin y$ and see what you got (like the other answer on here) :) $\endgroup$ – Chinny84 Aug 5 '15 at 16:11
  • $\begingroup$ Hope I will also get such an X-Ray vision, because being taught and then teaching is my plan for life) $\endgroup$ – blitzar787 Aug 5 '15 at 16:21
  • $\begingroup$ Correct. However, he has to be aware that multiplying by $cos(y)$ he may have multiplied by zero... So one has to be careful when establishing the interval where $y$ (as a solution) is defined, since it may be much shorter than the one you would get simply solving the equation after the multiplication by $cos(y)$... $\endgroup$ – bartgol Aug 5 '15 at 19:33
4
$\begingroup$

$$z=\sin(y) \implies \frac{dz}{dx}=\cos(y) \cdot \frac{dy}{dx}$$

So, in your equation you get:

$$\cos(y) \cdot \frac{dy}{dx} + \sin(y) = \frac{dz}{dx} + z = x$$

$\endgroup$

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.