# Math Help - Two questions

1. ## Two questions

I have some more questions.
dy/dx = 1 + x - y - xy, solve the differential equation.

I've been thinking how I can separate the variables before doing the integration, but I could not figure out how because of the xy. Please give me some hint to start! TQ =)

(cosx)^2* dy/dx = y
i'm stucked at integral of du / 1+ cos u = integral of dy / y. with 2x substituted as u.
I do not know how to proceed from there.

2. Originally Posted by custer
I have some more questions.
dy/dx = 1 + x - y - xy, solve the differential equation.

I've been thinking how I can separate the variables before doing the integration, but I could not figure out how because of the xy. Please give me some hint to start! TQ =)

[snip]
The differential equation re-arranges into $\frac{dy}{dx} + (x + 1) y = x + 1$.

This is solved using the integrating factor method.

Originally Posted by custer
[snip]
(cosx)^2* dy/dx = y
i'm stucked at integral of du / 1+ cos u = integral of dy / y. with 2x substituted as u.
I do not know how to proceed from there.
The differential equation re-arranges into $\int \! \frac{1}{y} \, dy = \int \! \sec^2 x \, dx$.

Both integrals are trivial.

3. sorry but I do not see how you get (sec x)^2

4. Originally Posted by custer
sorry but I do not see how you get (sec x)^2
By definition, $\frac{1}{\cos^2 x} = \sec^2 x$. I wrote $\sec^2 x$ rather than $\frac{1}{\cos^2 x}$ because the former is how the derivative of $\tan x$ is usually stated ....