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« **on:** January 26, 2018, 01:18:34 PM »
**Question**

Find the general solution of the given function.

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

**Answer**

Let y = xv,

$\frac{dy}{dx} = v + v'x$

$v+v'x = -\frac{4x+3xv}{2x+xv}$

$v+v'x = -\frac{(4+3v)}{(2+v)}$

$v'x = \frac{-v^2-5v-4}{2+v}$

$\int \frac{2+v}{(v+1)(v+4)}dv = \int \frac{1}{x} dx $

Let u = $(v^2+5v+4)$, du = (2v+5)dv

Then,

$\int \frac{2+v}{(v+1)(v+4)}dv = \int \frac{1}{2} \frac{(2v+5)}{(v^2+5v+4)}dv - \int \frac{0.5}{(v^2+5v+4)}dv$

$\frac{1}{2} ln(v^2+5v+4)dv - \frac{1}{2}*\frac{1}{3} \int(\frac{1}{v+1}-\frac{1}{v+4})dv$

$3ln(v^2+5v+4)-ln(\frac{(v+1)}{(v+4)}) = c-6ln(x)$

$(v+1)^2(v+1)^4 = x^6$

Therefore, we can get to know that

$(v+4)^2(v+1) = \frac{c}{x^{3}}$

plug y = xv into this equation

Therefore, we can get

$$(4x+y)^{2} (x+y) = c$$