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# Differential Equations C4 watch

1. Any ideas on how to solve?
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2. (Original post by ReubenOrion)
Any ideas on how to solve?
It's a separable equation. You can separate it into

and hence integrate both sides.

Seeing as this is a 'hence' question, you should be able to use part (a) to help with this.
3. (Original post by ReubenOrion)
Any ideas on how to solve?
well you separate the variables. once the y bits are on the left you will need to use partial fractions to integrate.

on the right you will have 1/cotx which can be expressed in a different way. they are giving you a clue as to the method of integration by mentioning sec2

the 2 can either stay on the left or move to the right as 1/2.
4. (Original post by the bear)
well you separate the variables. once the y bits are on the left you will need to use partial fractions to integrate.

on the right you will have 1/cotx which can be expressed in a different way. they are giving you a clue as to the method of integration by mentioning sec2

the 2 can either stay on the left or move to the right as 1/2.
I have integrated the partial fraction and 1/2 tanx

and have gotten

*ln 2+y / 2-y = 1/2 ln sec x + ln k

After substitution I have gotten

ln k = -ln sqrt2

I then tried to put this back into * and solve for sec x^2 however I am unsure where I have gone wrong as the answer is sec^2x = 8+4y/ 2-y
I have integrated the partial fraction and 1/2 tanx

and have gotten

*ln 2+y / 2-y = 1/2 ln sec x + ln k
Doesn't look right. The LHS is missing a 1/4 in front of the ln. The RHS seems OK. The constant term looks OK as well.

Once you see why you need the 1/4 there, you can mult through by 4 and hence turns into
6. so you get

½∫ 1/{2 + y} + 1/{2 - y} dy ={ln|secx|} + c

====> ln √ {(2 + y)/(2 - y)} = {ln|secx|} + c

now find c using the information given....

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