Differential equations (ODEs)

Original post by DFranklin

No (the fact the denominator has different powers of x and y stops it being homogeneous).

Any tips? Bit stuck.

Reply 3

Original post by jordanwu

Any tips? Bit stuck.

You can divide top and bottom by

x^3

to obtain

f(\frac{y}{x})

on the RHS

Reply 4

Original post by RDKGames

You can divide top and bottom by

x^3

to obtain

f(\frac{y}{x})

on the RHS

dy/dx = (x^2 + (2y^3 / x)) / y^2 ?

Reply 5

Original post by jordanwu

dy/dx = (x^2 + (2y^3 / x)) / y^2 ?

Not quite. In other words, divide each term by

x^3

Reply 6

Original post by RDKGames

Not quite. In other words, divide each term by

x^3

Each term? (x^3 / x^3 + 2y^3 / x^3) / (xy^2 / x^3) is what I did

Reply 7

Original post by jordanwu

Each term? (x^3 / x^3 + 2y^3 / x^3) / (xy^2 / x^3) is what I did

Yeah. Simplify some of those fractions to come to a clear expression that is

f(\frac{y}{x})

, and use an obvious substitution.

Reply 8

DFranklin

Er, seeing the subsequent posts, I think I'm wrong and it is homogeneous - sorry!

Reply 9

Original post by RDKGames

Yeah. Simplify some of those fractions to come to a clear expression that is

f(\frac{y}{x})

, and use an obvious substitution.

dy/dx = (x^2 / y^2) + 2y / x ?

Reply 10

Original post by jordanwu

dy/dx = (x^2 / y^2) + 2y / x ?

No need to split them. Have you covered the substitution

y(x)=xu(x)

Reply 11

Original post by RDKGames

No need to split them. Have you covered the substitution

y(x)=xu(x)

No I don't think I have

Reply 12

Original post by jordanwu

No I don't think I have

Well then you can't really do it without the substitution

\displaystyle u=\frac{y}{x}

Reply 13

Original post by RDKGames

Well then you can't really do it without the substitution

\displaystyle u=\frac{y}{x}

Is it y= xv, v= y/x or something? Got it in my notes on homogeneous ODEs

Reply 14

Original post by jordanwu

Is it y= xv, v= y/x or something? Got it in my notes on homogeneous ODEs

Then yes you have covered it... use it!

Reply 15

Original post by RDKGames

Then yes you have covered it... use it!

When you said not to split it did you mean I should keep it as one fraction? As in dy/dx = (x^2 / y^2)(1 + (2y^3 / x^3)) ?

Reply 16

Original post by jordanwu

When you said not to split it did you mean I should keep it as one fraction? As in dy/dx = (x^2 / y^2)(1 + (2y^3 / x^3)) ?

Keep it as

\dfrac{1+2\cdot \frac{y^3}{x^3}}{\frac{y^2}{x^2}}

as then after the conversion to

u

you'll want the reciprocal of this.

Reply 17

Original post by RDKGames

Keep it as

\dfrac{1+2\cdot \frac{y^3}{x^3}}{\frac{y^2}{x^2}}

as then after the conversion to

u

you'll want the reciprocal of this.

Is the substitution v= y^2 / x^2 ?

Reply 18

Original post by jordanwu

Is the substitution v= y^2 / x^2 ?

No. You have it in your notes, just follow those?

Reply 19