Question Number 208264 by efronzo1 last updated on 09/Jun/24
$$\:\:\:\cancel{\underline{\underbrace{ }}} \\ $$
Answered by mr W last updated on 10/Jun/24
![let y′=p p′−2p=2(e^x cos x−1) p=((2∫e^(−2x) (e^x cos x−1)dx+C_1 )/e^(−2x) ) =((e^(−x) (sin x−cos x)+e^(−2x) +C_1 )/e^(−2x) ) =e^x (sin x−cos x)+1+C_1 e^(2x) y=∫[e^x (sin x−cos x)+1+C_1 e^(2x) ]dx ⇒y=−e^x cos x+x+C_1 e^(2x) +C_2](https://www.tinkutara.com/question/Q208273.png)
$${let}\:{y}'={p} \\ $$$${p}'−\mathrm{2}{p}=\mathrm{2}\left({e}^{{x}} \mathrm{cos}\:{x}−\mathrm{1}\right) \\ $$$${p}=\frac{\mathrm{2}\int{e}^{−\mathrm{2}{x}} \left({e}^{{x}} \mathrm{cos}\:{x}−\mathrm{1}\right){dx}+{C}_{\mathrm{1}} }{{e}^{−\mathrm{2}{x}} } \\ $$$$\:\:\:\:=\frac{{e}^{−{x}} \left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)+{e}^{−\mathrm{2}{x}} +{C}_{\mathrm{1}} }{{e}^{−\mathrm{2}{x}} } \\ $$$$\:\:\:\:={e}^{{x}} \left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)+\mathrm{1}+{C}_{\mathrm{1}} {e}^{\mathrm{2}{x}} \\ $$$$\:{y}=\int\left[{e}^{{x}} \left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)+\mathrm{1}+{C}_{\mathrm{1}} {e}^{\mathrm{2}{x}} \right]{dx} \\ $$$$\:\Rightarrow{y}=−{e}^{{x}} \mathrm{cos}\:{x}+{x}+{C}_{\mathrm{1}} {e}^{\mathrm{2}{x}} +{C}_{\mathrm{2}} \\ $$