C4 First Order ODE

baffled by a question from the old heinemann "revise for C4' books.

Solve the differential equation,

2tanx.dy/dx + y^2 = 1

given that y = 1/2 when x = pi/6

its from Q6 integration test yourself section in any of you have the book.

i can only seem to end up with y^2 = 1 - 3/2 . (sinx)

the actual answer is supposedly y = (6sinx - 1)/(6sinx + 1)

any ideas of how to end up with that??

thanks. any quick help much appreciated

Reply 1

rbnphlp

11

thespacedonkey

baffled by a question from the old heinemann "revise for C4' books.

Solve the differential equation,

2tanx.dy/dx + y^2 = 1

given that y = 1/2 when x = pi/6

its from Q6 integration test yourself section in any of you have the book.

i can only seem to end up with y^2 = 1 - 3/2 . (sinx)

the actual answer is supposedly y = (6sinx - 1)/(6sinx + 1)

any ideas of how to end up with that??

thanks. any quick help much appreciated

how did you end up with that?
rearrange for dy/dx and the bring all the terms with x to dx and y to dy and then integrate

\displaystyle\int\frac{1}{1-y^2}dy=\int\frac{1}{2tanx}dx

and then consider partial fraction for LHS and integrate

Reply 2

thespacedonkey

OP

^
yep i did that, and when i integrated i got

-ln(1 - y) - ln(1 + y) = 1/2 . ln(sinx) + c

Reply 3

thespacedonkey

OP

if thats right...

then its after that where i think i've messed up

this is what i've done next:

(1 - y)(1 + y) = -1/2 .ksinx

1 - y^2 = -1/2 .ksinx

y^2 = 1 + 1/2 .ksinx

sub in y = 0.5 and x = pi/6

1/4 = 1 + 1/2 .k .1/2

therefore k = -3

and y^2 = 1 -3/2 .sinx

Reply 4

rbnphlp

11

thespacedonkey

if thats right...

then its after that where i think i've messed up

this is what i've done next:

(1 - y)(1 + y) = -1/2 .ksinx

1 - y^2 = -1/2 .ksinx

y^2 = 1 + 1/2 .ksinx

sub in y = 0.5 and x = pi/6

1/4 = 1 + 1/2 .k .1/2

therefore k = -3

and y^2 = 1 -3/2 .sinx