# Differential Equations: (x - 2y - 1) dy = (2x - 4y - 5) dx

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Differential Equations: (x - 2y - 1) dy = (2x - 4y - 5) dx

Coefficient Linear in Two Variables? Help please.

(x-2y-1)dy=(2x-4y-5)dx

By inspection, we can see that both sides of the equation contains x - 2y.
$(x - 2y - 1)\,dy = (2x - 4y - 5)\,dx$

Let
z = x - 2y
dz = dx - 2dy
dx = dz + 2dy

$(z - 1)\,dy = (2z - 5)(dz + 2dy)$

$(z - 1)\,dy = (2z - 5)\,dz + 2(2z - 5)\,dy$

$(z - 1)\,dy - 2(2z - 5)\,dy = (2z - 5)\,dz$

$(z - 1 - 4z + 10)\,dy = (2z - 5)\,dz$

$(9 - 3z)\,dy = (2z - 5)\,dz$

$dy = \dfrac{(2z - 5)\,dz}{9 - 3z}$

$dy = \left( -\dfrac{2}{3} + \dfrac{1}{9 - 3z} \right) \, dz$

$y + c = -\frac{2}{3}z - \frac{1}{3}\ln (9 - 3z)$

$y + c = -\frac{2}{3}(x - 2y) - \frac{1}{3}\ln \Big[ 9 - 3(x - 2y) \Big]$

$-\frac{1}{3}y + c = -\frac{2}{3}x - \frac{1}{3}\ln (9 - 3x + 6y)$

$y + c = 2x + \ln (9 - 3x + 6y)$       answer

Can we factor out (9-3x+6y) to make it 3ln(3-x+2y) is that valid?

Note that ln (xy) = ln (x) + ln (y), the same as
ln (ax) = ln (a) + ln (x) ≠ a ln (x).

For
ln (9-3x+6y) = ln [3(3-x+2y)] = ln (3) + ln (3-x+2y)

Oh. Thank you, sir.

how did you do it? dy=(2z−5)dz9−3z
dy=(2z−5)dz9−3z

long division