differential equations solutions by substitution please help =(?

Question

im doing solutions by substitutions now and im wondering how do u know when to u use y=ux and x=vy, or can u use either and get the same answer??
thanlks!!
ie on this one they used x=vy
xdx +(y-2x)dy=0

okay what, i dont understand Im sorry!! =|
please help my test is in less than 4 hours!!

Answer 1

xdx +(y-2x)dy=0. Put x = vy, so that dx = vdy + y dv. Then,

x( vdy + y dv) + (y -2x) dy =0

xv dy + xy dv + ydy - 2x dy =0

v^2y dy + v y^2 dv + ydy - 2vydy =0 Since y is a common factor,

v^2 dy + vydv + dy - 2vdy = 0

(v^2 - 2v+1)dy + vydv =0 Dividing by v,

(v -2 +1/v)dy + ydv =0 => (v -2 +1/v)dy = -ydv, now a difr. equation with separable variables.

-dy/dy = dv( v -2 +1/v). Now, we have to integrate both members. Int dy/y = ln(y). To integrate the 2nd,

Int vdv/(v^2 - 2v +1) = Int v/(v-1)^2 dv = Int (v-1)/(v-1)^2 dv + Int 1/(v-1)^2 dv = Int 1/(v-1) dv + Int 1/(v-1)^2 dv = ln(v-1) - 1/(v-1) + C, the integration constant.

So, y = exp(-(ln(v-1) - 1/(v-1) + C)). and v = x/y This is the solution, now it's algebra.

It'd be possible to put y = ux, but I think it'd be more complicated

Answer 2

See when xdy+ydx comes after substitution then according to that u substitute either y=ux or x=vy

Answer 3

Given: dy/dx = y/x + ?(y/x) it particularly is a Bernouilli equation. it may be much less perplexing to place in writing it like so: y' ? y/x = ?(y/x) = ?y/?x Divide the two facets by potential of two?y. the explanation why we do this could desire to be sparkling in the subsequent step. y'/(2?y) ? ?y/(2x) = a million/(2?x) enable u = ?y. observe that u' = y'/(2?y). u' ? u/(2x) = a million/(2?x) we've a universal first-order basic differential equation. enable ?(x) = e^(? [-a million/(2x)] dx) = a million/?x, and multiply the two facets by potential of ?: u'/?x ? u/(2?x³) = a million/(2x) observe that -a million/(2?x³) is the by-fabricated from a million/?x. we are in a position to as a effect replace like so: u'/?x + u * d/dx [a million/?x] = a million/(2x) by potential of the product rule, it particularly is comparable to: (u/?x)' = a million/(2x) combine the two facets: u/?x = ? [a million/(2x)] dx = ½ln(x) + C Multiply the two facets by potential of ?x: u = ?x[½ln(x) + C] replace back in the fee of u: ?y = ?x[½ln(x) + C] sq. the two facets: y = x/4 [ln(x) + C]²

Answer 4

The trick is to get it to the simplest possible form to inetgrate to get the answer.. so try whichever subsitution reduces power in the denominator etc; There is no hard and fast rule.. it varies according to the problem.