+9673 Solve the differential equation:
\(2xy\text{ dx} + (1+x^2)\text{ dy} = 0\)
Hi Max
separate variables to get
(1+x^2)dy = -2xydx
1/y dy = (-2x dx)/(1+x^2)
intergrate lhs wrt y and rhs wrt to x
ln y = - ln (1+x^2) = ln(1/{1+x^2} )
anti log both sides
y= 1/(1+x^2)
and a happy new year
+9673 OK. I thought it was correct...... Let me do it myself......
\(2xy\text{ dx} + (1+x^2) \text{ dy} = 0\\ 2xy\text{ dx} = -(1+x^2) \text{ dy}\\ -\dfrac{2x}{1+x^2} \text{ dx} = \dfrac{\text{ dy}}{y}\\ \displaystyle\int -\dfrac{2x}{1+x^2} \text{ dx} = \displaystyle\int\dfrac{\text{ dy}}{y}\\ \text{ Let }u = 1 + x^2,\text{ du} = 2x \text{ dx}\\ -\displaystyle\int \dfrac{\text{ du}}{u} = \displaystyle\int\dfrac{\text{ dy}}{y}\\ -\ln |u| = \ln |y| + C\\ \ln |\dfrac{1}{u}| = \ln |y| + C\\ \ln |\dfrac{1}{1+x^2}|= \ln |y| + C\\ \text{ Set }C = \ln K\\ \ln (\dfrac{1}{1+x^2})=\ln |y + K|\\ |y + K| = \dfrac{1}{1+x^2}\\ y + K = \pm \dfrac{1}{1+x^2}\\ y = K\pm\dfrac{1}{1+x^2}\)
