Answer:
[tex]-[ln(x^2-yx-y^2)] = K\\[/tex]
Step-by-step explanation:
Given the differential equation [tex](x+2y)y'=2x-y[/tex], this can also be written as;
[tex](x+2y)\frac{dy}{dx} =2x-y[/tex]
On simplification
[tex](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}[/tex]
The differential equation becomes;
[tex]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}[/tex]
[tex]\frac{dx}{x} = \frac{1+2v}{2-2v-2v^2}dv\\\\integrating\ both \ sides\\\\[/tex]
[tex]\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)[/tex]
[tex]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\\[/tex]
Hence the solution to the differential equation is [tex]-[ln(x^2-yx-y^2)] = K\\[/tex]
