Split - Trigonometric DE

[NOORALHUDA]

kindly i need help in this
initial value problem
x[cos][/2] (y) dx + tan y dy = 0 , y(1)=4

 

[Deleted member 4993]

kindly i need help in this
initial value problem
x[cos][/2] (y) dx + tan y dy = 0 , y(1)=4

As written - does not make sense to me. Please review your post - and may be post a photocopy of the assignment as it was given to you.

Please show us what you have tried and exactly where you are stuck.​

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[HallsofIvy]

kindly i need help in this
initial value problem
x[cos][/2] (y) dx + tan y dy = 0 , y(1)=4

I have no idea what "[cos][/2](y)" means. Here are a couple of guesses:
you meant "[^2]" or second power:
x cos^2(y)dx+ tan(y) dy= 0.
x cos^2(y)dx= -tan(y)dy= [sin(y)/cos(y)]dy

''Separate variables"- divide both sides by cos^(x).

x dx= [sin(y)/cos^3(y)]dy.

Integrating x dx on the left gives (1/2)x^2+ C where C is an arbitrary constant.

To integrate the right side, let u= cos(y). Then du= -sin(y)dy, -du= sin(y) dy and we have -du/u^3= -u^{-3}du. Integrating, (1/2)u^{-2}+ C' where C' is another arbitrary constant.

So (1/2)x^2+ C= (1/2)(cos(x))^{-2}+ C'= (1/2)(1/cos^2(x))+ C'

We can multiply both sides by 2 then add combine the constants to get x^2= 1/cos^2(y)+ C''.

Another possibility is that your "cos[/2](y)" is simply "cos(y)/2".

Then your equation is (1/2)x cos(y)dx+ tan(y)dy= 0. Separating variables as before, xdx= 2 (sin(y)/cos^2(y))dy. Integrate as before. The only difference is that we have cos^2(y) rather than cos^3(y).