Question 3

The differential equation is $\frac{dy}{dx} = \frac{y^3(e^{5x} + 1)}{e^x(1 + y^4)}$. Separating variables: $\frac{1 + y^4}{y^3} dy = \frac{e^{5x} + 1}{e^x} dx \implies (y + y^{-3}) dy = (e^{4x} + e^{-x}) dx$. Integrating both sides: $\int (y + y^{-3}) dy = \int (e^{4x} + e^{-x}) dx \implies \frac{y^2}{2} - \frac{1}{2y^2} = \frac{e^{4x}}{4} - e^{-x} + C$. Given $y(0) = \frac{1}{\sqrt{2}}$. Substituting $x=0$: $\frac{1/2}{2} - \frac{1}{2(1/2)} = \frac{1}{4} - 1 + C \implies \frac{1}{4} - 1 = -\frac{3}{4} + C \implies C = 0$. The equation is $\frac{y^2}{2} - \frac{1}{2y^2} = \frac{e^{4x}}{4} - e^{-x}$. To find $y(\log_e 2)$, substitute $x = \log_e 2$, so $e^x = 2$ and $e^{4x} = 16$. $\frac{y^2}{2} - \frac{1}{2y^2} = \frac{16}{4} - \frac{1}{2} = 4 - 0.5 = 3.5 = \frac{7}{2}$. $y^2 - \frac{1}{y^2} = 7 \implies y^4 - 7y^2 - 1 = 0$. Solving for $y^2$: $y^2 = \frac{7 \pm \sqrt{49 + 4}}{2} = \frac{7 \pm \sqrt{53}}{2}$. Since $y^2 > 0$ and $y^2 - \frac{1}{y^2} = 7 > 0$, we must have $y^2 > 1$. Thus, $y^2 = \frac{7 + \sqrt{53}}{2}$. $y = \sqrt{\frac{7 + \sqrt{53}}{2}}$.