Question

Find the general solution of the following differential equation. Primes denote derivatives with respect to x.(x+2y)y'=2x-yleft parenthesis x plus 2 y right parenthesis y prime equals 2 x minus y

Answer 1

Answer:

$-[ln(x^2-yx-y^2)] = K$

Step-by-step explanation:

Given the differential equation $(x+2y)y'=2x-y$, this can also be written as;

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

On simplification

$(x+2y)\frac{dy}{dx} =2x-y\\\\\frac{dy}{dx} = \frac{2x-y}{x+2y} \\\\let \ y = vx\\\frac{dy}{dx} = v+x\frac{dv}{dx}$

The differential equation becomes;

$v+x\frac{dv}{dx} =\frac{ 2x-vx}{x+2vx}\\\\v+x\frac{dv}{dx} = \frac{ x(2-v)}{x(1+2v)}\\\\v+x\frac{dv}{dx} = \frac{2-v}{1+2v}\\\\x\frac{dv}{dx} = \frac{2-v}{1+2v} - v\\\\x\frac{dv}{dx} = \frac{(2-v)-v(1+2v)}{1+2v}\\\\x\frac{dv}{dx} = \frac{2-v-v-2v^2}{1+2v}\\\\x\frac{dv}{dx} = \frac{2-2v-2v^2}{1+2v}$

$\frac{dx}{x} = \frac{1+2v}{2-2v-2v^2}dv\\\\integrating\ both \ sides$

$\int\limits \frac{dx}{x} = \int\limits \frac{1+2v}{2-2v-2v^2}dv\\\\lnx = \frac{1}{2} \int\limits \frac{1+2v}{1-v-v^2}dv\\\\lnx + C = -\frac{1}{2}ln(1-v-v^2)$

$C = -\frac{1}{2}ln(1-v-v^2) - lnx \\\\ -ln(1-v-v^2) - 2lnx = 2C\\\\-[ln(1-v-v^2) + lnx^2] = 2C\\\\-[ln(1-v-v^2)x^2] = 2C\\since\ v = y/x\\\\- [ln(1-y/x-y^2/x^2)x^2] = K\\\\-[ln(x^2-yx-y^2)] = K$

Hence the solution to the differential equation is $-[ln(x^2-yx-y^2)] = K$