# Thread: Differential Equations Intergating Factors

1. ## Differential Equations Intergating Factors

Hi

i have a few question i am having trouble solving:
Could someone tell me where my mistakes are.
1) $\displaystyle \frac{dy}{dx}+\frac{y}{x}=y^3$

$\displaystyle \frac{dy}{dx}+y=xy^3$

$\displaystyle \frac{dy}{dx}-y^{-2}=\frac{-1}{x}$

$\displaystyle I=e^{-x}$

$\displaystyle e^{-x}(\frac{dy}{dx}-y^2)=\frac{e^{-x}}{x}$

$\displaystyle e^{-x}y=\frac{e^{-x}}{x}$

what should i do next if this is correct.

2) $\displaystyle xcosy\frac{dy}{dx}-siny=0$

someone tell me how to start this?

i tried to integrate from this:
$\displaystyle \frac{dy}{dx} -tany = \frac{1}{x}$

but this didn't work.
P.S

2. With your first question, you managed to make three mistakes in a row. This problem seems to me to be Bernoulli. I think you need to use that procedure before doing the integrating factors method (which I think you WILL need when you get yourself a first-order LINEAR equation).

With your second equation, again, you're making algebra mistakes.

You've got $\displaystyle x\cos(y)\,y'-\sin(y)=0.$ Dividing through by $\displaystyle \cos(y)$ yields $\displaystyle x\,y'-\tan(y)=0.$ Then what?

3. For 2) the DE is separable...

$\displaystyle x\cos{y}\,\frac{dy}{dx} - \sin{y} = 0$

$\displaystyle x\,\frac{dy}{dx} - \frac{\sin{y}}{\cos{y}} = 0$

$\displaystyle x\,\frac{dy}{dx} = \frac{\sin{y}}{\cos{y}}$

$\displaystyle \frac{1}{x}\,\frac{dx}{dy} = \frac{\cos{y}}{\sin{y}}$

$\displaystyle \int{\frac{1}{x}\,\frac{dx}{dy}\,dy} = \int{\frac{\cos{y}}{\sin{y}}\,dy}$

$\displaystyle \int{\frac{1}{x}\,dx} = \int{\frac{1}{u}\,du}$ after making the substitution $\displaystyle u = \sin{y}$

$\displaystyle \ln{|x|} + C_1 = \ln{|u|} + C_2$

$\displaystyle \ln{|x|} = \ln{|\sin{y}|} + C$ where $\displaystyle C = C_2 - C_1$

$\displaystyle \ln{|x|} -\ln{|\sin{y}|} = C$

$\displaystyle \ln{\left|\frac{x}{\sin{y}}\right|} = C$

$\displaystyle \left|\frac{x}{\sin{y}}\right| = e^C$

$\displaystyle \frac{x}{\sin{y}} = \pm e^C$

$\displaystyle \frac{x}{\sin{y}} = A$ where $\displaystyle A = \pm e^C$

$\displaystyle x = A\sin{y}$

$\displaystyle \frac{1}{A}x = \sin{y}$

$\displaystyle Bx = \sin{y}$ where $\displaystyle B = \frac{1}{A}$

$\displaystyle y = \arcsin{Bx}$.

4. oh i c, thank you

5. You're very welcome for whatever I was able to contribute. Have a good one!