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Question:

Consider the equation $y' + (\cos x)y = e^{-\sin(x)}$. Show that any solution $\varphi$ has the property that $\varphi(\pi k) - \varphi(0) = \pi k, k \in \mathbb{Z}$. General solution: $\varphi(x) = \frac{x}{e^{\sin(x)}} + \frac{c}{e^{\sin(x)}}$

I know this property is true but I do not know how to prove this. Please help me with this.

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    If you know what $\varphi$ is, what are you stuck on? – 2017-02-15
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    @anomaly what should I do next? – 2017-02-15
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    @anomaly Should I just substitute? $φ(πk)−φ(0) = (\frac{πk}{e^{sin(πk)}} + \frac{c}{e^{sin(πk)}}) - (0 + c)$ equal to what? – 2017-02-15
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    Yes you need to do that only. – 2017-02-15

1 Answers 1

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Well, you have the general solution:

$$\text{y}'\left(x\right)+\cos\left(x\right)\cdot\text{y}\left(x\right)=\exp\left(-\sin\left(x\right)\right)\space\Longleftrightarrow\space\text{y}\left(x\right)=\frac{x+\text{C}}{\exp\left(\sin\left(x\right)\right)}\tag1$$

So, now:

  • When $x=0$: $$\text{y}\left(0\right)=\frac{0+\text{C}}{\exp\left(\sin\left(0\right)\right)}=\frac{\text{C}}{\exp\left(0\right)}=\frac{\text{C}}{1}=\text{C}\tag2$$
  • When $x=\text{k}\pi$: $$\text{y}\left(\text{k}\pi\right)=\frac{\text{k}\pi+\text{C}}{\exp\left(\sin\left(\text{k}\pi\right)\right)}=\frac{\text{k}\pi+\text{C}}{\exp\left(0\right)}=\frac{\text{k}\pi+\text{C}}{1}=\text{k}\pi+\text{C}\tag3$$ Where $\text{k}\in\mathbb{Z}$ and using that $\sin\left(\text{k}\pi\right)=0$

So:

$$\text{y}\left(\text{k}\pi\right)-\text{y}\left(0\right)=\text{k}\pi+\text{C}-\text{C}=\text{k}\pi+0=\text{k}\pi\tag4$$

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    How is exp(sin(kπ)) = 0? – 2017-02-15
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    @socrates that's not what I stated in my answer! – 2017-02-15
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    @socrates It's not - but $\sin(k\pi)=0$, so $e^{\sin(k\pi)}=1$. – 2017-02-15
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    @StevenStadnicki how is $sin(kπ)=0$? – 2017-02-15
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    @socrates Well, the period of the sine function ($\sin\left(x\right)$) is $2\pi$ and the period of $\sin\left(\text{k}\pi\right)$ is $2$ (when $\text{k}\in\mathbb{Z}$) – 2017-02-15
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    @JanEerland If that equals to 2 then, it must be $e^2$, not $e^0$? – 2017-02-15
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    @socrates You do not understand what I say, I'm talking about the period not the value of the function, because the value of the function is equal to $0$ so we get indeed: $e^0=1$. So why the downvote? – 2017-02-15
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    @JanEerland I did not downvote – 2017-02-15
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    @socrates These are very, _very_ basic and fundamental properties of the sin function that we're talking about. Not to be too blunt about it, but how did you even come to the problem you're working on without having these under your belt? – 2017-02-15