1. ## Separable Differential Equation

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?

2. Originally Posted by littlesohi
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

3. 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

$\displaystyle y=\sqrt{13e^{2t}-4}$

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 $\displaystyle A$

$\displaystyle 3 = \sqrt{Ae^{2\times 0 }-4} = \dots$

5. 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 $\displaystyle y = \pm \sqrt{13e^{2x} - 4}$.