Solve the following differential equation and write the answer in the least simplified form.

𝑦′ + 𝑐𝑜𝑡(𝑥)𝑦 = 𝑠𝑖𝑛(𝑥) 

Solution:

To solve the differential equation 𝑦′ + 𝑐𝑜𝑡(𝑥)𝑦 = 𝑠𝑖𝑛(𝑥), we can use an integrating factor. The integrating factor is given by the exponential of the integral of the coefficient of 𝑦, which in this case is 𝑐𝑜𝑡(𝑥).

The integrating factor, `e^{∫cot(x)}dx`, can be written as `e^sin(x)`. Multiplying the entire differential equation by this integrating factor, we get:

`e^sin(x)y' + e^sin(x)cot(x)y = e^sin(x)sin(x)`

Now, we can rewrite the left-hand side of the equation as the derivative of the product of the integrating factor and 𝑦:

`(e^sin(x)y)' = e^sin(x)sin(x)`

Integrating both sides with respect to 𝑥, we have:

`∫(e^sin(x)y)′ dx = ∫e^sin(x)sin(x) dx`

`e^sin(x)y =  ∫e^sin(x)sin(x) dx`

Now, we can solve the integral on the right-hand side:

`e^sin(x)y = ∫sin^2(x) dx = (-1/2)cos(2x) + x/2`

Finally, dividing both sides by `e^sin(x)`, we get the solution for the differential equation:

`y = {(-1/2)cos(2x) + x/2} / e^sin(x)`