Question
Find the general solution of this de (1+cosx)y'=sinx(sinx+sinxcosx-y)
I need full step have tried but nothing good is coming out
I need full step have tried but nothing good is coming out
Answers
oobleck
Rearrange things into the usual form for a 1st-order linear DE:
(1+cosx)y' = sin^2x(1+cosx) - sinx y
y' - sinx/(1+cosx) y = sin^2x
y' - tan(x/2) y = sin^2x
Now you have the form y' + p(x) y = q(x)
The integrating factor is now e^∫p(x) dx = 1/2 sec^2(x/2)
1/2 sec^2(x/2) y' - tan(x/2)(1/2 sec^2(x/2))y = sin^2(x) * 1/2 sec^2(x/2)
After some trig manipulation, this comes out to be
(tan(x/2) y)' = 4(1+cos^2(x))/(1+cos(x))
Some more study should provide the way to integrate the right side.
Check my math, and don't be afraid to use tools like wolframalpha to provide some help.
(1+cosx)y' = sin^2x(1+cosx) - sinx y
y' - sinx/(1+cosx) y = sin^2x
y' - tan(x/2) y = sin^2x
Now you have the form y' + p(x) y = q(x)
The integrating factor is now e^∫p(x) dx = 1/2 sec^2(x/2)
1/2 sec^2(x/2) y' - tan(x/2)(1/2 sec^2(x/2))y = sin^2(x) * 1/2 sec^2(x/2)
After some trig manipulation, this comes out to be
(tan(x/2) y)' = 4(1+cos^2(x))/(1+cos(x))
Some more study should provide the way to integrate the right side.
Check my math, and don't be afraid to use tools like wolframalpha to provide some help.