Show that any solution $\varphi$ of $y' + (\cos x)y = e^{-\sin(x)}$ has the property that $\varphi(\pi k) - \varphi(0) = \pi k, k \in \mathbb {Z}$

Question

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.

Answer 1

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:

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