solve the given differential equation by using an appropriate substitution. the de is homogeneous…

Explanation:

Step1: Rewrite the differential equation

Given ((y^{2}+yx)dx - x^{2}dy = 0), we can rewrite it as (\frac{dy}{dx}=\frac{y^{2}+yx}{x^{2}}). Let (y = vx), then (\frac{dy}{dx}=v + x\frac{dv}{dx}).

Step2: Substitute (y = vx) into the differential equation

Substituting (y = vx) into (\frac{dy}{dx}=\frac{y^{2}+yx}{x^{2}}), we get (v + x\frac{dv}{dx}=\frac{(vx)^{2}+(vx)x}{x^{2}}). Simplify the right - hand side: (\frac{v^{2}x^{2}+vx^{2}}{x^{2}}=v^{2}+v). So the equation becomes (v + x\frac{dv}{dx}=v^{2}+v).

Step3: Solve the resulting differential equation for (v)

Subtract (v) from both sides of (v + x\frac{dv}{dx}=v^{2}+v), we obtain (x\frac{dv}{dx}=v^{2}). Separate the variables: (\frac{dv}{v^{2}}=\frac{dx}{x}).

Step4: Integrate both sides

Integrate (\int\frac{dv}{v^{2}}=\int\frac{dx}{x}). The left - hand side integral is (\int v^{- 2}dv=-\frac{1}{v}+C_1), and the right - hand side integral is (\int\frac{dx}{x}=\ln|x|+C_2). So (-\frac{1}{v}=\ln|x| + C) (where (C = C_2 - C_1)).

Step5: Substitute back (v=\frac{y}{x})

Substitute (v=\frac{y}{x}) into (-\frac{1}{v}=\ln|x| + C), we get (-\frac{x}{y}=\ln|x|+C), or (y =-\frac{x}{\ln|x| + C}).

Answer:

(y =-\frac{x}{\ln|x|+C})