Question

\(\left(x^{2} y+x^{2}\right) d x+\left(y^{2} x-y^{2}\right) d y=0\)

Short answer

Question: Solve the first-order homogeneous differential equation \(\left(x^{2} y+x^{2}\right) d x+\left(y^{2} x-y^{2}\right) d y=0\). Answer: The solution to the homogeneous differential equation is given by the equation \(y = \pm \frac{x}{x^2+1} \left( C - x^2 \right)\), where C is the constant of integration.

Step-by-step solution

Recognize the equation type and substitution method

We are given a first-order differential equation: \(\left(x^{2} y+x^{2}\right) d x+\left(y^{2} x-y^{2}\right) d y=0\). Since it is homogeneous, we can use the substitution method by introducing a new variable \(v = \frac{y}{x}\).

Recast the equation in terms of v

We know that \(v = \frac{y}{x}\), so we can rewrite \(y\) as \(y = xv\). Then differentiate with respect to \(x\) to get \(\frac{dy}{dx} = v + x\frac{dv}{dx}\). Replace \(y\) and \(\frac{dy}{dx}\) in the original differential equation.

Substitute and simplify the equation

Substitute \(y = xv\) and \(\frac{dy}{dx} = v + x\frac{dv}{dx}\) into the original equation and then simplify: \((x^2(xv) + x^2)dx + ((xv)^2x - (xv)^2)dy = 0\) \((x^3v + x^2)dx + (x^3v^2 - x^2v^2)(v + x\frac{dv}{dx})dy = 0\) Now, notice that this differential equation is in the form of an exact differential equation. Hence, we can rewrite it as: \(\frac{d}{dx}(x^3v + x^2) + (x^3v^2 - x^2v^2)(v + x\frac{dv}{dx}) = 0\)

Solve for the integral

Integrate both sides of the equation with respect to x: \(\int \left(\frac{d}{dx}(x^3v + x^2) + (x^3v^2 - x^2v^2)(v + x\frac{dv}{dx})\right) dx = \int 0 dx\) On the left-hand side, integration with respect to x is the same as taking the antiderivative of the given function. So, integrating and simplifying, we get: \(x^3v + x^2 + x^3v^3 - x^2v^3 = C\) Now, let's substitute back the original variable \(y\). Since \(v = \frac{y}{x}\), we have: \(x^3\left(\frac{y}{x}\right) + x^2 + x^3\left(\frac{y}{x}\right)^3 - x^2\left(\frac{y}{x}\right)^3 = C\)

Final solution

Simplify the equation to obtain the final solution: \(y = \pm \frac{x}{x^2+1} \left( C - x^2 \right)\)