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# Solving Differential Equations by substitution watch

1. For ln((1+x)/(1-x)) it would need to be ln(1+x) - ln(1-x) but I'm not getting B = minus (number)

Where am I messing up?
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2. you should get ln |u| = -ln |1 - x2| + c for part i) then make u the subject.

your problem is the integral of 1/{2{ 1 - x } }... it is -1/2*ln { 1 - x }, not + 1/2*ln { 1 - x }
3. (Original post by the bear)
you should get ln |u| = -ln |1 - x2| + c for part i) then make u the subject.

your problem is the integral of 1/{2{ 1 - x } }... it is -1/2*ln { 1 - x }, not + 1/2*ln { 1 - x }
u = A /1-x^2 is the solution it says u = dy/dx

so dy/dx = A / 1-x^2 I meant to find the solution from that I don't get how
4. (Original post by ckfeister)
u = A /1-x^2 is the solution it says u = dy/dx

so dy/dx = A / 1-x^2 I meant to find the solution from that I don't get how
all of your working is fine up until the integral of 1/{2{1-x}} where you missed off the minus sign.
5. (Original post by the bear)
all of your working is fine up until the integral of 1/{2{1-x}} where you missed off the minus sign.
6. if you integrate 1/( 1 - x ) you get - ln | 1 - x | NOT ln | 1 - x |

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