- #1
- 436
- 13
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.
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.
