# What kind of ODE is this and how to solve it?

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

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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 $\mathrm{d}v = z\mathrm{d}u + u \mathrm{d}z$:

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

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I didn't see your answer. However, I explained the motivation of the method to solve the equation so I guess there is no problem. – Pedro Tamaroff Feb 26 '12 at 16:53
@Peter no worries! It is good that you explained the general method for finding the change of coordinates. – Willie Wong Feb 26 '12 at 17:04

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 $\dfrac{Y}{X} = v$

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