0
$\begingroup$

I want to know which can be a particular integral for this equation: $$y' - 2y = \cos x.$$ I have these possible solutions:

  1. $y = \frac{1}{5} \sin x - \frac{2}{5} \cos x$
  2. $y = -\frac{1}{5} \sin x - \frac{2}{5} \cos x$
  3. $y = e^{2x}$
  4. $y = \sin x$

I thought it can be 2. but I am not sure about it. I hope someone can reply.

Edit: After your answers, this is what I've tried:

For 2.

$y' = \frac{1}{5} \sin{x} -\frac{2}{5} \cos{x}$

$y = -\frac{1}{5} \sin{x} -\frac{2}{5} \cos{x}$

$- \frac{1}{5} \cos{x} + \frac{2}{5} \sin{x} - 2 (-\frac{1}{5} \sin{x} - \frac{2}{5} \cos{x}) = \cos{x}$

$- \frac{1}{5} \cos{x} + \frac{2}{5} \sin{x} + \frac{2}{5} \sin{x} + \frac{4}{5} \cos{x} \neq \cos{x}$

Therefore, it cannot be 2.

For 3.

The 3. cant' be because:

$y' =e^{2x}$

$y= e^{2x}$

in this way the equation:

$ e^{2x} - 2 e^{2x} \neq \cos{x}$

For the 1.

$y' = \frac{1}{5} \cos{x} +\frac{2}{5} \sin{x}$

$y = \frac{1}{5} \sin{x} -\frac{2}{5} \cos{x}$

$ \frac{1}{5} \cos{x} +\frac{2}{5} \sin{x} - 2( \frac{1}{5} \sin{x} -\frac{2}{5} \cos{x} ) = cos{x} $

$ \frac{1}{5} \cos{x} + \frac{2}{5} \sin{x} - \frac{2}{5} \sin{x} +\frac{4}{5} \cos{x} = cos{x} $

m.c.m within $\frac{1}{5}$ and $\frac{4}{5}$ and i obtain $\frac{5}{5}$ that is just 1. So i obtain $cos{x}$ = $cos{x}$

what do you think?

  • 4
    What have you tried? Where have you gotten stuck? It's rude to just ask people to do your homework for you (especially after having been told so after [asking](http://mathoverflow.net/questions/262891/particular-integral-from-one-equation) this question [twice](http://mathoverflow.net/questions/262948/) already). – 2017-02-23
  • 0
    i want only some Hint to understand the work. @NoahSchweber – 2017-02-23
  • 2
    Well, what have you tried, and where did you get stuck? – 2017-02-23

6 Answers 6

4

Hint -

You have linear equation of the form $\frac{dy}{dx} + Py =Q$

Find integrating factor $e^{\int P dx}$ then multiply it with equation. Then integrate.

Solution -

I want you to learn steps. So I solved it for you. Hope it helps.

$y' - 2y = \cos x$

I.F = $e^{-2x}$

So above equation becomes,

$y'.e^{-2x} - 2y.e^{-2x} = \cos x.e^{-2x}$

$\frac{d}{dx} (y.e^{-2x}) = \cos x.e^{-2x}$

Integrating both sides,

$y.e^{-2x} = \int \cos x.e^{-2x} dx$ .....(1)

Now let $I = \int \cos x.e^{-2x} dx$

On integration by parts taking $e^{-2x}$ as first term,

$I = e^{-2x}. \sin x + 2 \int e^{-2x}. \sin x dx$

Again by parts on last term taking $e^{-2x}$ as first term,,

$I = e^{-2x}. \sin x - 2 e^{-2x}. \cos x -4 \int e^{-2x}. \cos x dx $

$I = e^{-2x}. \sin x - 2 e^{-2x}. \cos x - 4I $

$5I = e^{-2x}. \sin x - 2 e^{-2x}. \cos x$

$I = \frac{e^{-2x}. \sin x - 2 e^{-2x}. \cos x}{5}$

Putting in equation (1),

$y.e^{-2x} = \frac{e^{-2x}. \sin x - 2 e^{-2x}. \cos x}{5} + c$

$y = \frac{\sin x - 2 \cos x}{5} + ce^{2x}$

  • 0
    Any doubt feel free to ask. – 2017-02-23
  • 0
    i wil try thank you @Amar not everypeople are gently like you here. – 2017-02-23
  • 0
    So nice of you. If I know how to solve I am always ready to explain. – 2017-02-23
  • 0
    i am trying with the 1. if you have some other hint i am happy to hear it thank you Amar – 2017-02-23
  • 0
    Bro I edit my answer. If any doubt let me know. Hope you are boy. Lol. – 2017-02-23
  • 0
    Bro thank you, i am a boy yes XD, i am so happy someone is gently to help me like you, but now i am little bit confused, i must only understand if the sum of the derivate and normal y is equals to cosx or i must integrate too? – 2017-02-23
  • 0
    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/54137/discussion-between-amar-and-amiga-500). – 2017-02-23
  • 0
    @Amiga500 This is just another way to approach the problem (by solving the general solution). You can also do it with my method, where you substitute $y$ and $y'$ for each possible solution. – 2017-02-23
4

Hint: in each of the possible answers, what's the value of $y' - 2y$?

1

Since you are just testing whether they can be particular solutions, it suffices to just substitute each of the possible solutions on $y'-2y$, and verifying if that value is equal to $\cos{x}$.

For example, for number (4): $$y=\sin{x}$$ $$y'=\cos{x}$$ Substituting gives: $$\cos{x}-2\sin{x}\not\equiv \cos{x}$$ Therefore, (4) is not a solution.

Try this with the other possible solutions, and you will find that one of them works.

1

Cheat by computing $y'-2y$ for the given proposals

  1. $\frac{1}{5} \sin x - \frac{2}{5} \cos x\to\frac{1}{5} \cos x + \frac{2}{5} \sin x-\frac{2}{5} \sin x + \frac{4}{5} \cos x$
  2. $ -\frac{1}{5} \sin x - \frac{2}{5} \cos x\to-\frac{1}{5} \cos x + \frac{2}{5} \sin x+\frac{2}{5} \sin x +\frac{4}{5} \cos x$
  3. $e^{2x}\to 2e^{2x}-2e^{2x}$
  4. $\sin x\to\cos x-2\sin x$

So 1. (You don't even have to compute the others.)

0

The 3. cant' be because:

$y' =e^{2x}$

$y= e^{2x}$

in this way the equation:

$ e^{2x} - 2 e^{2x} != cosx$

what do you think?

  • 0
    the first, yes you can of course – 2017-02-23
  • 1
    Your derivative is wrong - $y'$ is $2e^{2x}$. Remember the chain rule! $e^{2x}$ is the function "$2x$" sitting inside the function "$e^x$", so the derivative is $e^{2x}\cdot 2$ (or better, $2e^{2x}$) - the derivative of the outside, applied to the inside, times the derivative of the inside. That said, your conclusion - that 3. is wrong - is correct. – 2017-02-23
  • 0
    thank you @projectilemotion i will remember it, and so the 3. is not the right solution. The 1. and the 2. can be. – 2017-02-23
  • 0
    thank you to @NoahSchweber too, now i must only try with the 1. and the 2. – 2017-02-23
  • 0
    @Amiga500 Only one of the solutions are correct, it is either option (1) or (2). Could you mention in the comments the derivatives you get for each of them ($y'$)? – 2017-02-23
  • 0
    i am just trying it and i post it very soon@projectilemotion – 2017-02-23
0

The 2.

$y' = 1/5 sinx -2/5 cosx$

$y = -1/5sinx -2/5 cosx$

$- 1/5 cosx + 2/5 sinx - 2 (-1/5 sinx - 2/5 cosx) = cosx$

$- 1/5 cosx + 2/5 sinx + 2/5 sinx + 4/5 cosx != cosx$

what do you think?

  • 0
    Good try, however you've made a mistake with finding the derivative $y'$. Since the derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin{x}$, your derivative should be: $$y'=-\frac{1}{5}\cos{x}+\frac{2}{5}\sin{x}$$ – 2017-02-23
  • 0
    By the way, you should rather add your attempts like this on your question instead as of an answer. I will add your attempt on your question if you don't mind. – 2017-02-23
  • 0
    sorry me @projectilemotion i will do it with the last number 1. Yes you can do it of course – 2017-02-23
  • 0
    No problem. I upvoted your question since you made some good effort. Since you found that (2), (3), (4) is wrong, you should find that (1) is correct assuming you do (1) correctly. – 2017-02-23