Answer to Question #105859 in Differential Equations for Pappu Kumar Gupta

Question

solve D.E. y''=1+(y' )^2

Answer 1

$y'' = 1 + (y')^2\\ \frac{dy'}{dx} = 1 + (y')^2\\ \frac{dy'}{1 + (y')^2} = dx\\ x = \intop\frac{dy'}{1 + (y')^2} = arctan(y') + c_1\\ y' = tan(x - c_1)\\ y = \intop tan(x - c_1)dx = \intop \frac{sin(x - c_1)}{cos(x - c_1)}d(x - c_1) = \intop - \frac{dcos(x - c_1)}{cos(x - c_1)} = - ln|cos(x - c_1)| + c_2$