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

 

[blue_raver22]

Hello,

The general solution to this differential equation is y(x) = -2cos(x) + C. To understand why this is the case, we can first rewrite the equation as:

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

Then, we can use the trigonometric identity \sin(2x) = 2\sin(x)\cos(x) to rewrite the right side as:

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

Next, we can recognize that \frac{\sin(x)}{\cos(x)} is simply the tangent function, so we can rewrite the equation as:

y' = 2\tan(x)\cos(x)

Now, we can use the chain rule to find the derivative of y with respect to x:

y' = \frac{dy}{dx} = \frac{dy}{d\tan(x)}\frac{d\tan(x)}{dx} = \sec^2(x) \cdot 2\tan(x)

Substituting this back into our original equation, we get:

\frac{dy}{dx} = 2\tan(x)\cos(x) = 2\frac{\sin(x)}{\cos(x)}\cos(x) = 2\sin(x)

Integrating both sides with respect to x, we get:

y = \int 2\sin(x)dx = -2\cos(x) + C

So the general solution to the differential equation is y(x) = -2cos(x) + C. I hope this helps clarify the method used to solve this problem.