Finding an implicit, general solution to a homogeonous differential equation

JNBirDy
Messages
37
Reaction score
0

Homework Statement


Find an implicit, general solution to:

dy/dx = (6x - 4y) / (x - y) with x > 0.


Homework Equations





The Attempt at a Solution



dy/dx = (6x - 4y) / (x - y)

x(dv/dx) + v = (6 - 4v) / (1 - v)

[(1-v) / (v-3)(v-2)] dv = dx / x

\int dv/(v-2) - 2\int dv/(v-3) = \int dx/x

ln|v-2| -2ln|v-3| + C_{1} = ln|x| + C_{2}

ln|v-2| -2ln|v-3|- ln|x| = C_{3}

----

Not sure where to go from here, I'm trying to get it into the form

|y - ax| / (y - bx)^c = C, for some constants a, b, c, and C.

----

Any help is appreciated, thanks.
 
Physics news on Phys.org
JNBirDy said:
ln|v-2| -2ln|v-3|- ln|x| = C

The integration looks fine. Don't bother bringing everything to one side: you want to leave the function of x on the right, thus ln|v-2| - 2 ln|v-3| = ln|x| + C . Use the properties of logarithms to write the left-hand side as the logarithm of a single expression, then "exponentiate" both sides to eliminate the logarithms. You will have a function x(v) , from which point you could "back-substitute" for v . (I haven't checked yet to see whether this will be nicely-enough-behaved algebraically to get y(x) .)

EDIT: It isn't, but I see that you were only asked to find an implicit solution involving x and y . That would make sense: the absolute values in the result for v indicate that the solution is a curve which "fails the Vertical Line Test", so it must be split into parts, each of which is the curve for a function.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top