Question Number 190851 by Rupesh123 last updated on 13/Apr/23
Answered by mr W last updated on 13/Apr/23
![cos y′=sin y sin ((π/2)−y′)=sin y (π/2)−y′=nπ+(−1)^n y −y′=(−1)^n y+(((2n−1)π)/2) −(dy/dx)=(−1)^n y+(((2n−1)π)/2) (dy/((−1)^n y+(((2n−1)π)/2)))=−dx ∫(dy/((−1)^n y+(((2n−1)π)/2)))=−∫dx ln [(−1)^n y+(((2n−1)π)/2)]=−(−1)^n x+C_1 (−1)^n y+(((2n−1)π)/2)=Ce^((−1)^(n+1) x) ⇒y=(−1)^n [Ce^((−1)^(n+1) x) −(((2n−1)π)/2)], n∈Z](https://www.tinkutara.com/question/Q190865.png)
$$\mathrm{cos}\:{y}'=\mathrm{sin}\:{y} \\ $$$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−{y}'\right)=\mathrm{sin}\:{y} \\ $$$$\frac{\pi}{\mathrm{2}}−{y}'={n}\pi+\left(−\mathrm{1}\right)^{{n}} {y} \\ $$$$−{y}'=\left(−\mathrm{1}\right)^{{n}} {y}+\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}} \\ $$$$−\frac{{dy}}{{dx}}=\left(−\mathrm{1}\right)^{{n}} {y}+\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}} \\ $$$$\frac{{dy}}{\left(−\mathrm{1}\right)^{{n}} {y}+\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}}}=−{dx} \\ $$$$\int\frac{{dy}}{\left(−\mathrm{1}\right)^{{n}} {y}+\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}}}=−\int{dx} \\ $$$$\mathrm{ln}\:\left[\left(−\mathrm{1}\right)^{{n}} {y}+\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right]=−\left(−\mathrm{1}\right)^{{n}} {x}+{C}_{\mathrm{1}} \\ $$$$\left(−\mathrm{1}\right)^{{n}} {y}+\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}}={Ce}^{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}} \\ $$$$\Rightarrow{y}=\left(−\mathrm{1}\right)^{{n}} \left[{Ce}^{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}} −\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right],\:{n}\in\mathbb{Z} \\ $$
Commented by Rupesh123 last updated on 13/Apr/23
Excellent, sir!