Determine the general solution to the differential equation
y′ =$\displaystyle \frac{y}{x^2+4x+5}$
Fix the constant of integration according to the initial condition y(−1) = 1
$\displaystyle \frac{dy}{dx} = \frac{y}{x^2+4x+5}$
$\displaystyle \frac{dy}{y} = \frac{dx}{x^2+4x+5}$
$\displaystyle \frac{dy}{y} = \frac{dx}{(x+2)^2+1}$
Remember that $\displaystyle \int \frac{dx}{x^2+a^2} = \frac{1}{a}\arctan(\frac{x}{a}) + C $, and $\displaystyle \int \frac{dx}{x} = \ln{|x|} + C $
$\displaystyle \frac{dy}{dx} = \frac{y}{x^2+4x+5}$
$\displaystyle \frac{dy}{y} = \frac{dx}{(x+2)^2 + 1}$
$\displaystyle \ln|y| = \arctan(x+2) + C$
$\displaystyle \ln(1) = \arctan(-1+2) + C$
$\displaystyle 0 = \arctan(1) + C$
$\displaystyle C = -\frac{\pi}{4} $
finish up the solution