# [SOLVED] Differential Equation

• Jan 25th 2009, 03:45 PM
ronaldo_07
[SOLVED] Differential Equation
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
• Jan 25th 2009, 03:48 PM
Mush
Quote:

Originally Posted by ronaldo_07
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 $\displaystyle \int \frac{dx}{x} = \ln{|x|} + C$
• Jan 25th 2009, 03:51 PM
skeeter
Quote:

Originally Posted by ronaldo_07
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

$\displaystyle \frac{dy}{dx} = \frac{y}{x^2+4x+5}$

$\displaystyle \frac{dy}{y} = \frac{dx}{(x+2)^2 + 1}$

$\displaystyle \ln|y| = \arctan(x+2) + C$

$\displaystyle \ln(1) = \arctan(-1+2) + C$

$\displaystyle 0 = \arctan(1) + C$

$\displaystyle C = -\frac{\pi}{4}$

finish up the solution
• Jan 25th 2009, 04:23 PM
ronaldo_07
to get y= shal I just take log on both sides so y=$\displaystyle e^{arctan}$......
• Jan 25th 2009, 04:33 PM
skeeter
$\displaystyle \ln|y| = \arctan(x+2) - \frac{\pi}{4}$

change to an exponential equation ...

$\displaystyle y = e^{\arctan(x+2) - \frac{\pi}{4}}$