Answer to Question #140452 in Differential Equations for Anurag Dhayal

Question

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

Answer 1

"\\displaystyle\n\ny'' + (y')^2 = 2e^{-y}\\\\\n\n\\textsf{Let}\\, y = ln(f(x))\\\\\n\ny' = \\frac{f'(x)}{f(x)}\\\\\n\ny'' = \\frac{f(x)f"(x) - f'(x)\u00b2}{f(x)\u00b2}\\\\\n\ny'' + (y')^2 = 2e^{-y} \\implies \\frac{f(x)f"(x) - f'(x)^2}{f(x)\u00b2} + \\left(\\frac{f'(x)}{f(x)}\\right)^2 = \\frac{2}{f(x)} \\\\\n\n\\frac{f(x)f"(x)}{f(x)\u00b2} - \\frac{2}{f(x)} = 0\\\\\n\n\n\\frac{f"(x)}{f(x)} - \\frac{2}{f(x)} = 0\\\\\n\n\n\\implies f"(x) - 2 = 0\\\\\n\n\nf"(x) = 2\\\\\n\n\\implies \\int f"(x)\\, \\mathrm{d}x = \\int 2 \\, \\mathrm{d}x\\\\\n\nf'(x) = 2x + C_1\\\\\n\n\n\\int f'(x)\\, \\mathrm{d}x = \\int 2x + C_1\\, \\mathrm{d}x\\\\\n\nf(x) = x^2 + C_1x + C_2\\\\\n\n\n\\therefore y = \\ln(x^2 + C_1x + C_2)\\\\\\textsf{is the general solution}\\\\\n\\textsf{to the differential equation, where}\\\\ C_1 \\, \\textsf{and}\\, C_2 \\textsf{are arbitrary constants.}"