Simple Differential Equation: Finding General Solution for y'\cos(x) = \sin(2x)

[FeDeX_LaTeX]

Hello;

Found an exercise on simple differential equations on some website, got all correct except for this one. It only supplies answers but no method, but am stuck as to how they got their answer. Asked to find a general solution to the following differential equation:

$$y'\cos(x) = \sin(2x)$$

Here's my method:

Had to make it in the form y' = f(x), so;

$$y' = \frac{\sin(2x)}{\cos(x)}$$

Integrating both sides gives us;

$$y = \int \frac{\sin(2x)}{\cos(x)}dx$$

EDIT: Forget about method I wrote underneath. I saw my error. But can anyone show me how/why the above equates to -2cos(x) + C?

Thanks.

 

[tiny-tim]

Hello FeDeX_LaTeX!

You really need to learn your trigonometric identities …

in this case, sin2x = 2sinxcosx ​

 

[FeDeX_LaTeX]

Ack! I completely missed that! I knew that identity and it just completely fell out of my head... haha. Thanks! :)

 

[terryphi]

see attached