1. ## differential equations

Hello!

this is my homework:

y' = sin(1x)/2y

can you help me please i'm in a hurry and have no idea how to solve this! I tried solving it like a homogenous DE but I didn't know how to transform DE into y/x form.

regards, Vide

EDIT: y' is derivate of y

2. As it's written, it's just a separable ODE.

3. ODE?
If i want to solve it like a homogenous DE i need to transform SINx/2y in a
y/x form so i can write u=y/x but i don't know how to transform it. Or is it DE of some other type, not homogenous?

4. i tried basic separable metod and the result is (-cosx)^1/2 so i think that is homogenous DE.

5. Originally Posted by Vide
Hello!

this is my homework:

y' = sin(1x)/2y

can you help me please i'm in a hurry and have no idea how to solve this! I tried solving it like a homogenous DE but I didn't know how to transform DE into y/x form.

regards, Vide

EDIT: y' is derivate of y
$\displaystyle \frac{dy}{dx}=\frac{\sin(x)}{2y}\Rightarrow{2ydy=\ sin(x)dx}$

integrating we get

$\displaystyle \int{2ydy}=y^2=\int\sin(x)dx=-\cos(x)+C$

solving for y we get

$\displaystyle y=\pm\sqrt{C-\cos(x)}$

6. thanks! I forgot C now i see

7. Originally Posted by Vide
thanks! I forgot C now i see
Haha...yeah its ok...I have done that once or twice

8. Originally Posted by Mathstud28
$\displaystyle \frac{dy}{dx}=\frac{\sin(x)}{2y}\Rightarrow{2ydy=\ sin(x)dx}$

integrating we get

$\displaystyle \int{2ydy}=y^2=\int\sin(x)dx=-\cos(x)+C$

solving for y we get

$\displaystyle y=\pm\sqrt{C-\cos(x)}$
When solving a DE, we don't (well I don't) enjoy seeing $\displaystyle \pm$ in the solution. Leaving it as the implicit solution $\displaystyle y^2=C-\cos(x)$ would be more preferable. Your answer would be acceptable, given certain restricitions on y.