Question

form the differential equation for the following family of curves y=c1*(e^x)+(e^c1)*e^(2x)+(ln(c1))*e^(3x) .

Answer 1

Question #16055

Given, $y(x) = C_1 e^x + e^{C_1} e^{2x} + \ln C_1 e^{3x}$ , lets make substitution $e^x = t$ . So, $y(t) = C_1 t + e^{C_1} t^2 + \ln C_1 t^3$ . Then, find the derivatives:

$y'(t) = 3t^2\ln C_1 + 2e^{C_1}t + C_1; y''(t) = 6t\ln C_1 + 2e^{C_1}; y'''(t) = 6\ln C_1 \Rightarrow C_1 = e^{\frac{y'''}{6}}$ . So, the differential equation is $y(t) = \frac{y'''}{6} t^3 + \exp(\exp(\frac{y'''}{6})) t^2 + \exp(\frac{y'''}{6}) t$ .