first order linear differential equations involving sin and cos?

Question

can somebody show me how to solve this please.
cos(x) dy/dx + y sin(x) = 1

Answer 1

well, first divide by cosx, to get
y'+tan(x)y=secx. This is linear and can be solved by finding the integrating factor Mu, which i will call M.
M=E^(integral of (tan(x))), the integral of tangent is found by doing a u substitution. Since tan=sin/cos, let u=cos, so du=
-sinxdx. Thus you get -integral (1/u)du. which is -ln(cos(x)). So M=E^-ln(cosx).
I hope you know how to do simpler linear diff eqs, cause I am going to skip a step. After multiplying both sides by M and integrating the left side, you get
E^-ln(cosx)*y=Integral of [-(ln(cosx)secx)dx]

I'm pretty sure it's impossible to integrat the right hand side, so leave it as and integral, and divide by E^-ln(cosx) to get what y is. Sorry I couldn't get a definitive answer.

Answer 2

If you let y= A sin x + B cos x
Then y' = A cos x - B sin x
Substituting, A cos^2 x - B cos x sin x +
A sin^2 x - B cos x sin x = 1
However, cos^2 x + sin^2 x = 1, so A = 1 and B=0
Then y= sin x

Answer 3

let y=sin(x). then dy/dx = cos(x), and the equation is satisfied.

if you want the general solution, then you also need to solve the homogeneous equation
cos(x) dy/dx + y sin(x) = 0,
which can be rewritten as
dy/dx = -y tan(x).

you can use separation of variables:
1/y dy = -tan(x) dx

then integrate:
ln|y| = ln|sec(x)| + C

and solve for y:
y = Ae^(sec(x)),
where A is a constant (it arises when we pull out e^C in the earlier equation, but unlike e^C it can be negative, because we also removed the absolute value signs).

so the general solution to the equation you gave is:
y = sin(x) + Ae^(sec(x)).