Question

solve the given differential equation by using an appropriate substitution. the de is homogeneous. ((y^{2}+yx)dx - x^{2}dy = 0)

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:

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