• Aug 26th 2006, 03:03 PM
simfonija
(a) 1/x * dy/dx + 2/(1+x^2)*y=3 y(1)=4

(b) 2*x * dy/dx = 2*y + (x^2+y^2)^1/2

Thank you very much
• Aug 26th 2006, 04:09 PM
galactus
I'll give you the solution to b, you try to work to that end. Okey-doke?.

b): $\displaystyle \frac{5}{x^{2}+1}+\frac{3(x^{2}+1)}{4}$
• Aug 26th 2006, 04:49 PM
ThePerfectHacker
Quote:

Originally Posted by simfonija
(a) 1/x * dy/dx + 2/(1+x^2)*y=3 y(1)=4

$\displaystyle \frac{1}{x}\frac{dy}{dx}+\frac{2}{1+x^2}y=3$
Multiply by "x",
$\displaystyle \frac{dy}{dx}+\frac{2x}{1+x^2}y=3x$
This is a Linear-First Order Differential Equation.
Here,
$\displaystyle P(x)=\frac{2x}{1+x^2}$
$\displaystyle Q(x)=3x$
Therefore, $\displaystyle u=1+x^2$
$\displaystyle I(x)=\exp \left( \int \frac{2x}{1+x^2} dx\right) =e^{\ln (1+x^2)}=1+x^2$
Thus, the solutions are,
$\displaystyle \frac{1}{I(x)}\cdot \int I(x)Q(x) dx$
Thus,
$\displaystyle (1+x^2)^{-1}\int \frac{3x}{1+x^2} dx$
Use, $\displaystyle u=1+x^2$,
$\displaystyle (1+x^2)^{-1}(3/2\ln (1+x^2)+C)$
Now you probably mean,
$\displaystyle y(0)=4$ because these are usually how initial conditions are given.
Thus,
$\displaystyle C=4$
From here the unique solution is,
$\displaystyle \frac{3\ln (1+x^2)}{2 (1+x^2)}+\frac{4}{1+x^2}$