Solving Linear First Order D.E. y' - y*tan[x] = 2sin[x]

[splitendz]

Solve y’ – y*tan[x] = 2sin[x].

I keep arriving at the answer: y = 2ln(sec[x]) / sec[x] - c/sec[x]
for this question.

According to my textbook the correct answer is: y = (c - cos[x]) * sec[x]. Can anyone explain how to obtain this answer?

Thanks :)

 

[dextercioby]

Show us your work.Because of that natural logarithm,the two answers are not equivalent (equal for some value of the integration constant),therefore at least one of them is wrong.

Daniel.

 

[HallsofIvy]

The text answer is correct. I suspect you overlooked the fact that integrating will give you a ln(y) as well as "ln(sec(x))".

Since this is a linear equation, you can do the homogeneous and non-homogeneous parts separately. The associated homogeneous equation is just y'- y tan x= 0 or
y'= y tan x so dy/y= tan x dx= (sin x dx)/cos x.
To find a solution to the non-homogeneous part you could use "variation of parameters": try a solutution of the form y(x)= u(x) sec(x) (I'm giving away part of the answer there!), differentiate and plug into the equation to find a simple equation for u(x).
Or just guess a simple solution!

 

[splitendz]

Ah right. Thanks a lot for your help HallsofIvy! I've got it now.