Solucionar d^2y/dx^2+9y=sec3x

$$\frac { d ^ { 2 } y } { d x ^ { 2 } } + 9 y = \sec 3 x$$

Resolver para x (solución compleja)

$$x=\frac{2}{3}i\ln(3)+\left(-\frac{1}{3}i\right)\ln(\left(1+\left(\left(-81\right)y^{2}+1\right)^{\frac{1}{2}}\right)y^{-1})+\frac{2}{3}\pi n_{8}\text{, }n_{8}\in \mathrm{Z}\text{, }\nexists n_{3}\in \mathrm{Z}\text{ : }\frac{2}{3}i\ln(3)+\left(-\frac{1}{3}i\right)\ln(\left(1+\left(\left(-81\right)y^{2}+1\right)^{\frac{1}{2}}\right)y^{-1})+\frac{2}{3}\pi n_{8}=\frac{1}{3}\pi n_{3}+\frac{1}{6}\pi \text{ and }\nexists n_{3}\in \mathrm{Z}\text{ : }\frac{2}{3}i\ln(3)+\left(-\frac{1}{3}i\right)\ln(\left(1+\left(\left(-81\right)y^{2}+1\right)^{\frac{1}{2}}\right)y^{-1})+\frac{2}{3}\pi n_{8}=\frac{1}{3}\pi n_{3}+\frac{1}{6}\pi$$

$$x=\left(-\frac{1}{3}i\right)\ln(\left(\frac{1}{9}+\left(-\frac{1}{9}\right)\left(\left(-81\right)y^{2}+1\right)^{\frac{1}{2}}\right)y^{-1})+\frac{2}{3}\pi n_{9}\text{, }n_{9}\in \mathrm{Z}\text{, }\nexists n_{3}\in \mathrm{Z}\text{ : }\left(-\frac{1}{3}i\right)\ln(\left(\frac{1}{9}+\left(-\frac{1}{9}\right)\left(\left(-81\right)y^{2}+1\right)^{\frac{1}{2}}\right)y^{-1})+\frac{2}{3}\pi n_{9}=\frac{1}{3}\pi n_{3}+\frac{1}{6}\pi \text{ and }\nexists n_{3}\in \mathrm{Z}\text{ : }\left(-\frac{1}{3}i\right)\ln(\left(\frac{1}{9}+\left(-\frac{1}{9}\right)\left(\left(-81\right)y^{2}+1\right)^{\frac{1}{2}}\right)y^{-1})+\frac{2}{3}\pi n_{9}=\frac{1}{3}\pi n_{3}+\frac{1}{6}\pi$$

Resolver para y (solución compleja)

$$y=\frac{1}{9\cos(3x)}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{3}+\frac{\pi }{6}$$

Resolver para y

$$y=\frac{1}{9\cos(x)\left(4\left(\cos(x)\right)^{2}-3\right)}$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(x>\frac{\pi n_{1}}{3}+\frac{\pi }{6}\text{ and }x<\frac{\pi n_{1}}{3}+\frac{\pi }{2}\right)$$