Question

y"+(y') ^2=2e^-y find the general solution

Answer 1

$\displaystyle y'' + (y')^2 = 2e^{-y}\\ \textsf{Let}\, y = ln(f(x))\\ y' = \frac{f'(x)}{f(x)}\\ y'' = \frac{f(x)f"(x) - f'(x)²}{f(x)²}\\ y'' + (y')^2 = 2e^{-y} \implies \frac{f(x)f"(x) - f'(x)^2}{f(x)²} + \left(\frac{f'(x)}{f(x)}\right)^2 = \frac{2}{f(x)} \\ \frac{f(x)f"(x)}{f(x)²} - \frac{2}{f(x)} = 0\\ \frac{f"(x)}{f(x)} - \frac{2}{f(x)} = 0\\ \implies f"(x) - 2 = 0\\ f"(x) = 2\\ \implies \int f"(x)\, \mathrm{d}x = \int 2 \, \mathrm{d}x\\ f'(x) = 2x + C_1\\ \int f'(x)\, \mathrm{d}x = \int 2x + C_1\, \mathrm{d}x\\ f(x) = x^2 + C_1x + C_2\\ \therefore y = \ln(x^2 + C_1x + C_2)\\\textsf{is the general solution}\\ \textsf{to the differential equation, where}\\ C_1 \, \textsf{and}\, C_2 \textsf{are arbitrary constants.}$