# Thread: exact DE (with integrating factor)

1. ## exact DE (with integrating factor)

Okay here's the problem, I can't seem to find an integrating factor for this differential equation. As the original equation doesn't come out to be exact, we need an integrating factor to make it exact then we can solve it as an exact DE.

The equation is;
[sin(y)cos(y) + xcos^2(y)]dx + x.dy = 0 ---(1)

which is the same as;
[1/2sin(2y) + xcos^2(y)]dx + x.dy = 0

now if we let;
A = 1/2sin(2y) + xcos^2(y)
and
B = x

and take the partial derivatives of both;

dA/dy = cos(2y) - xsin(2y)
dB/dx = 1

and since dA/dy does not equal dB/dx we use either

i. (1/A)[dB/dx - dA/dy]
or
ii. (1/B)[dA/dy - dB/dx]

to get a function dependant only on y or only on x

...and I can't get that. I've tried various trig identities with no success, maybe I've messed up the basic differentiation there, or maybe I've missed an identity, I'm not sure. If someone can help me with the integrating factor, I'm fairly certain I can find the general solution to the equation easily.

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1. using separation of variables, solve;
(1+2x)cosy .dx + dy/cosy = 0

for this one, I end up doing t-substitution partial fractions to integrate it :/ is that right?? Because I get;

INT [1+2x] dx = INT [-1/cos^2(y)] dy
= INT [-2/(cos(2y) +1)] dy

2. using the change of variable u=y^3, solve;
y' + yx^2 = [sinh(x) . e^(-x^3)]/3y^2

and this one I don't get very far on...I can't get rid of the dy/dx when i substitute in for u and du/dy. I get;

du/dy . dy/dx +3ux^2 = sinh(x). e^(-x^3)

and FYI: these are from a tutorial sheet, but they need to be done, and I'd much prefer to learn how to do them. As you can see, i've given all of them a crack.

2. Originally Posted by DistantCube
Okay here's the problem, I can't seem to find an integrating factor for this differential equation. As the original equation doesn't come out to be exact, we need an integrating factor to make it exact then we can solve it as an exact DE.

The equation is;
[sin(y)cos(y) + xcos^2(y)]dx + x.dy = 0 ---(1)

which is the same as;
[1/2sin(2y) + xcos^2(y)]dx + x.dy = 0

now if we let;
A = 1/2sin(2y) + xcos^2(y)
and
B = x

and take the partial derivatives of both;

dA/dy = cos(2y) - xsin(2y)
dB/dx = 1

and since dA/dy does not equal dB/dx we use either

i. (1/A)[dB/dx - dA/dy]
or
ii. (1/B)[dA/dy - dB/dx]

to get a function dependant only on y or only on x

...and I can't get that. I've tried various trig identities with no success, maybe I've messed up the basic differentiation there, or maybe I've missed an identity, I'm not sure. If someone can help me with the integrating factor, I'm fairly certain I can find the general solution to the equation easily.

.
I think you did everything correctly. It seems no integrating factor to turn the ODE into an exact one does not exist (at least not dependent on one variable).

1. using separation of variables, solve;
(1+2x)cosy .dx + dy/cosy = 0
You have, for some interval.

(1+2x)cos y + (1/cos y)y' = 0

Divide by cos y to get,

(1+2x) + sec^2 y * y' = 0

Integrate,

INT (1+2x) dx + INT sec^2 y *y' dx = C

x+x^2 + tan y =C

3. Heir.

4. Originally Posted by ThePerfectHacker
I think you did everything correctly. It seems no integrating factor to turn the ODE into an exact one does not exist (at least not dependent on one variable).
I found the integrating factor just now anyway. You can change the identities on the top to becomes;

[1-1+2sin^2y +2xsinycosy]/[sinycosy + xcos^2y]

then take out a factor of 2siny on the top, cosy on the bottom, the rest cancels and you're left with 2tany.

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edit: Also, for the question where you attached the image, I can't believe I didn't realise before - that's a Bernoulli equation! Could have simply used that form :/

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Thank you very much for your help, I shall work through these this afternoon. I shouldn't come across any trouble, but if I do, I'll know who to ask

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# differetial of( sinycosy xcos^2y)dx xdy=0

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