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$ (2x-4y+6)dx+(x+4y-3)dy=0 $ I converted it to this form $ dy/dx = (2x-4y+6)/(x+4y-3) $ in the hope I can use $ z=y/x $ substitution for homogeneous ODEs but because of constants 6 and 3 I can't.

2 Answers 2

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Do first a change of variables:

$ u = x + 1 \qquad v = y-1 $

Then $\mathrm{d}u = \mathrm{d}x$ and $\mathrm{d}v = \mathrm{d}y$. Now

$ 2u - 4v = 2x + 2 - 4y + 4 = 2x - 4y + 6$

and

$ u + 4v = x + 1 + 4y - 4 = x + 4y - 3$

So in the new variables you are solving

$ (2u-4v) \mathrm{d}u + (u+4v) \mathrm{d}v = 0 $

or (away from the line $u+4v = 0$)

$ \frac{2u-4v}{u+4v} \mathrm{d}u + \mathrm{d}v = 0 $

Then you can do the substitution: write $v = uz$ you get $\mathrm{d}v = z\mathrm{d}u + u \mathrm{d}z$:

$ \left( \frac{2-4z}{1+4z} + z\right) \mathrm{d}u + u \mathrm{d}z = 0 $

  • 0
    @Peter no worries! It is good that you explained the general method for finding the change of coordinates.2012-02-26
3

This is an ODE that can be solved in the following way. Write

$ (2x-4y+6)dx+(x+4y-3)dy=0 $

as

$\frac{-2x+4y-6}{x+4y-3} =\frac{dy}{dx}$

What we need now is to write the RHS as

$\frac{ax+by}{cx+dy} $

so that it becomes homogeneous.

So we need to find $k$ and $h$ such that

$-2(x_1+h)+4(y_1+k)-6=-2x_1 +4y_1$

$ (x_1+h)+4(y_1+k)-3 =x_1 +4y_1$

This is, we need that

$\eqalign{ & - 2h + 4k - 6 = 0 \cr & h + 4k - 3 = 0 \cr} $

This gives,

$\eqalign{ & h = - 1 \cr & k = 1 \cr} $

So we have

$\frac{{ - 2{x_1} + 4{y_1}}}{{{x_1} + 4{y_1}}} = \frac{{d{y_1}}}{{d{x_1}}}$

But this ODE is homogeneous (as we wanted), so we can put

$\frac{{{y_1}}}{{{x_1}}} = v$

and get

\frac{{ - 2 + 4v}}{{1 + 4v}} = v'{x_1} + v

This results in a separable ODE,

$\frac{{d{x_1}}}{{{x_1}}} = - \frac{{4v + 1}}{{4{v^2} - 3v + 2}}dv$

I guess you can take it from here.

NOTE: In general, the equation

$\frac{{ax + by + c}}{{dx + ey + f}} = \frac{{dy}}{{dx}}$

can be made homogeneous by solving

$\eqalign{ & ah + bk + c = 0 \cr & dh + ek + f = 0 \cr} $

and substituting $X = x+h$, $Y=y+k$.

In turn, the homogeneous ODE can be made separable by putting $\dfrac{Y}{X} = v$