I am having trouble with this equation:
$\displaystyle dy/dx = y + 4/y $
I multiplied by y to get a common denominator, but then got stucked there. Can anyone help me?
$\displaystyle \frac{dy}{dx} = y + \frac{4}{y} $
$\displaystyle \frac{dy} {y + \frac{4}{y}} = 1~dx $
$\displaystyle \frac{dy} { \frac{y^2+4}{y}} = 1~dx $
$\displaystyle \frac{y}{y^2+4} ~dy = 1~dx $
$\displaystyle \int \frac{y}{y^2+4} ~dy = \int 1~dx $
Use $\displaystyle u = y^2+4$ on the LHS
Okay I did that and got:
$\displaystyle ln y^2 + 4 = t + c $
took the power of e to cancel ln :
$\displaystyle y^2 + 4 = e ^{t+c} $
since this is an initial-value problem y(0) = 3 .. you solved for C and got that C is approx. 2.56 .. but my answer is not quite similar to the one of the back of the book which says
$\displaystyle y=\sqrt{13e^{2t}-4} $
$\displaystyle \int \frac{y}{y^2+4} ~dy = \int 1~dx$
$\displaystyle \frac{1}{2}\int \frac{2y}{y^2+4} ~dy = \int 1~dx$
$\displaystyle \frac{1}{2}\int \frac{du}{u} = x+C$
$\displaystyle \frac{1}{2}\ln u = x+C$
$\displaystyle \frac{1}{2}\ln (y^2+4) = x+C$
$\displaystyle \ln (y^2+4) = 2x+C$
$\displaystyle y^2+4 = e^{2x+C}$
$\displaystyle y^2+4 = e^{2x}e^{C}$
$\displaystyle y^2+4 = Ae^{2x}$
$\displaystyle y^2 = Ae^{2x}-4$
$\displaystyle y = \sqrt{Ae^{2x}-4}$
Now use $\displaystyle y(0) = 3$ to solve for $\displaystyle A$
$\displaystyle 3 = \sqrt{Ae^{2\times 0 }-4} = \dots$
Using alternative (and better) notation:
$\displaystyle \frac{dy}{dx} = y + \frac{4}{y}$
$\displaystyle \frac{dy}{dx} = \frac{y^2 + 4}{y}$
$\displaystyle \left(\frac{y}{y^2 + 4}\right)\frac{dy}{dx} = 1$
$\displaystyle \int{\left(\frac{y}{y^2 + 4}\right)\frac{dy}{dx}\,dx} = \int{1\,dx}$
$\displaystyle \int{\frac{y}{y^2 + 4}\,dy} = x + C_1$
$\displaystyle \frac{1}{2}\int{\left(\frac{1}{y^2 + 4}\right)2y\,dy} = x + C_1$
Now let $\displaystyle u = y^2 + 4$ so that $\displaystyle \frac{du}{dy} = 2y$
$\displaystyle \frac{1}{2}\int{\frac{1}{u}\,\frac{du}{dy}\,dy} = x + C_1$
$\displaystyle \frac{1}{2}\int{\frac{1}{u}\,du} = x + C_1$
$\displaystyle \frac{1}{2}\ln{|u|} + C_2 = x + C_1$
$\displaystyle \frac{1}{2}\ln{|y^2 + 4|} + C_2 = x + C_1$
$\displaystyle \frac{1}{2}\ln{(y^2 + 4)} = x + C_1 - C_2$ (notice we can get rid of the modulus because $\displaystyle y^2 + 4$ is always positive
$\displaystyle \ln{(y^2 + 4)} = 2x + C$, where $\displaystyle C = 2(C_1 - C_2)$
$\displaystyle y^2 + 4 = e^{2x + C}$
$\displaystyle y^2 + 4 = e^{2x}e^{C}$
$\displaystyle y^2 + 4 = Ae^{2x}$ where $\displaystyle A = e^{C}$
$\displaystyle y^2 = Ae^{2x} - 4$
$\displaystyle y = \pm \sqrt{Ae^{2x} - 4}$.
And now using your initial condition that $\displaystyle y(0) = 3$
$\displaystyle 3 = \pm \sqrt{Ae^{2(0)} - 4}$
$\displaystyle 3 = \pm \sqrt{A - 4}$
$\displaystyle 9= A - 4$
$\displaystyle A = 13$.
So finally, your solution is $\displaystyle y = \pm \sqrt{13e^{2x} - 4}$.