first order differentials

Hi could someone please help me?

The equation to solve is:

xdy/dx = y + (x^2 - y^2)^(1/2)

any ideas?
- it isn't in the form dy/dx + P(x)y=q(x) the main problem being the q(x) function.
- i have tried integrating factor approach to help separate it...but this is useless given q(x) has y terms in it.
- any suggested substitutions to use?
thanks!

Reply 1

Swayum

18

What level of maths are you at (A-level/1st year university/2nd year/etc)? I feel like there should be some sort of substitution given here. Are there any previous parts to the question?

Reply 2

ibush1

OP

hi well i'm doing sciences at uni (1st year) and this is the maths course for my course. i thought i could use sec(u) and tan(u) maybe?? that could make the
(x^2 - y^2)^(1/2) function disappear.

Reply 3

ibush1

OP

yes there are previous parts
the full part of the question is:

Solve the following first order differential equations:
i) x dy/dx = 3(1-y) given y=0 and x=1

ii) (1 + x^2)^(1/2)dy/dx=xexp(-y) given y=0 at x=0

iii) xdy/dx = y + (x^2 - y^2)^(1/2)

no boundary/initial conditions are given for iii).

Reply 4

ibush1

OP

it's fine i've got it now - thanks.

Reply 5

Swayum

18

Divide through by x first to get

dy/dx = y/x + ((x^2 - y^2)^0.5)/x

Then note that

\frac{(x^2 - y^2)^{\frac{1}{2}}}{x} = (\frac{x^2 - y^2}{x^2})^\frac{1}{2}

Now sub in u = y/x.

That may or may not help you. I haven't tried it, but I can't see anything else to do.

Reply 6

Farhan.Hanif93

18

Try

y=x\sin \theta

.

Reply 7

Farhan.Hanif93

18

Original post by ibush1

it's fine i've got it now - thanks.

What did you use to solve it?

Reply 8

anshul95

11

Original post by ibush1

Hi could someone please help me?

The equation to solve is:

xdy/dx = y + (x^2 - y^2)^(1/2)

any ideas?
- it isn't in the form dy/dx + P(x)y=q(x) the main problem being the q(x) function.
- i have tried integrating factor approach to help separate it...but this is useless given q(x) has y terms in it.
- any suggested substitutions to use?
thanks!

I used y=xcos u and it seems to reduce to a separable first order.

Reply 9

Swayum

18

Original post by Farhan.Hanif93

Try

y=x\sin \theta

.

Yeah, this is equivalent to y/x = u. The benefit of not putting sin(theta) in from the start is that the u substitution is a bit easier and the resulting integral can be recognized.

(edited 13 years ago)

Reply 10

Farhan.Hanif93

18

Original post by Swayum

Yeah, this is equivalent to y/x = u. The benefit of not putting sin(theta) in from the start is that the u substitution is a bit easier and the resulting integral can be recognized.

Fair enough.

first order differentials

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