Differential Equations

Concept:Solve a separable first-order differential equation by separating variables and integrating.Explanation:The given equation is $\frac{dy}{dx} = \frac{e^{5x}y^3 + y^3}{e^x + e^x y^4}$.Factor the right-hand side: $\frac{y^3(e^{5x}+1)}{e^x(1+y^4)}$.Separate variables: $\frac{1+y^4}{y^3}\,dy = \frac{e^{5x}+1}{e^x}\,dx$.Simplify the left: $\frac{1}{y^3} + y\,dy$. Simplify the right: $e^{4x} + e^{-x}\,dx$.Integrate both sides: $\int y^{-3}\,dy + \int y\,dy = \int e^{4x}\,dx + \int e^{-x}\,dx$.Result: $-\frac{1}{2y^2} + \frac{y^2}{2} = \frac{1}{4}e^{4x} - e^{-x} + C$.Multiply by $2$: $-\frac{1}{y^2} + y^2 = \frac{1}{2}e^{4x} - 2e^{-x} + 2C$.Use $y(0)=\frac{1}{\sqrt{2}}$ so $y^2(0)=\frac{1}{2}$. At $x=0$: left = $-2 + \frac{1}{2} = -\frac{3}{2}$, right = $\frac{1}{2} - 2 = -\frac{3}{2}$, so $C=0$.The equation becomes $y^2 - \frac{1}{y^2} = \frac{1}{2}e^{4x} - 2e^{-x}$.Let $u = y^2$. Then $u - \frac{1}{u} = \frac{1}{2}e^{4x} - 2e^{-x}$.Evaluate at $x = \log_e 2$: $e^{4x} = 16$, $e^{-x} = \frac{1}{2}$. So right side = $\frac{1}{2}\cdot16 - 2\cdot\frac{1}{2} = 8 - 1 = 7$.Thus $u - \frac{1}{u} = 7$ $\Rightarrow$ $u^2 - 7u - 1 = 0$.Solve: $u = \frac{7 \pm \sqrt{49+4}}{2} = \frac{7 \pm \sqrt{53}}{2}$.Since $y>0$, $u = y^2 > 0$, so $u = \frac{7 + \sqrt{53}}{2}$ (the other root is negative).Therefore $y = \sqrt{\frac{7 + \sqrt{53}}{2}}$.Answer:$\sqrt{\frac{7 + \sqrt{53}}{2}}$