0
$\begingroup$

Question:

Solve the following differential equation: $$\cos x dy = y(\sin x -y) dx$$

I simplified it down to: $$\frac{dy}{dx} = y\tan x - y^2\sec x$$

Not sure how I should proceed from here though. Any hints?

$\endgroup$
2
$\begingroup$

This is a Bernoulli equation.

$$y'=f(x)y+g(x)y^\alpha$$

Use $u=y^{1-\alpha}$ as substitution to reduce your problem to a linear ODE.

$$u'=(1-\alpha)(f(x)u+g(x))$$

$\endgroup$
  • $\begingroup$ Aha.... I see now. However, is there any other way to solve the problem? $\endgroup$ – Gummy bears Sep 16 '15 at 11:52
  • $\begingroup$ Yes there are but they are more general=harder to understand. $\endgroup$ – MrYouMath Sep 16 '15 at 11:53
  • $\begingroup$ The thing is I haven't been taught this.... So I'm not sure that this is the way that I should solve it. $\endgroup$ – Gummy bears Sep 16 '15 at 11:54
  • $\begingroup$ there might be other ways, but this is the quickest way to solve this. Maybe your teacher didn't select this equation properly. $\endgroup$ – MrYouMath Sep 16 '15 at 11:56
  • $\begingroup$ Ahh... might be. Thanks anyways. I guess it's good to learn new things. So after it becomes a linear ODE, we solve it the normal way, right? $\endgroup$ – Gummy bears Sep 16 '15 at 11:57
1
$\begingroup$

HINT:

Notice, $$\frac{dy}{dx}=y\tan x-y^2\sec x\iff \frac{dy}{dx}-y\tan x=-y^2\sec x$$

$$-\frac{1}{y^2}\frac{dy}{dx}+\frac{1}{y}\tan x=\sec x$$

Let $\frac{1}{y}=u\implies \frac{-1}{y^2}\frac{dy}{dx}=\frac{du}{dx}$, by substitution we get $$\color{red}{\frac{du}{dx}+u\tan x=\sec x}$$ I hope you can solve the above equation in Bernoulli's D.E. form.

$\endgroup$
  • $\begingroup$ Yeah that becomes a linear equation. That's the method I was looking for. $\endgroup$ – Gummy bears Sep 16 '15 at 12:09
1
$\begingroup$

Solution without $\bf{Bernoulli}$ Substution::

Given $$\cos xdy = y(\sin x-y)dx = y\sin xdx-y^2dx$$

So $$\displaystyle \cos xdy-y\sin xdx = -y^2 dx\Rightarrow \frac{\cos xdy-y\sin xdx}{y^2\cos^2 x} = -\frac{1}{\cos^2 x}dx$$

So we get $$\displaystyle d\left(\frac{1}{y\cos x}\right) = - \sec^2 xdx$$

$$\displaystyle \int \frac{d}{dx}\left(\frac{1}{y\cos x}\right)dx = - \int \sec^2 xdx$$

So we get $$\displaystyle \frac{1}{y\cos x} = -\tan x+\mathcal{C}$$

$\endgroup$
0
$\begingroup$

Hint: This is Bernoulli type ODE. To solve it, you need to make the change $z(x)=y(x)^{1-2}=\frac{1}{y(x)}$

$\endgroup$
  • $\begingroup$ Sorry but neither of those names help.... $\endgroup$ – Gummy bears Sep 16 '15 at 11:49
  • $\begingroup$ I edited my answer. $\endgroup$ – Svetoslav Sep 16 '15 at 11:50
  • $\begingroup$ Yeah...... What just happened? This is not something that I've been taught... at least not that I remember. $\endgroup$ – Gummy bears Sep 16 '15 at 11:50

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.