Question

The general solution of the differential equation \[ x^{3}\,dx-xy^{2}\,dy+y^{3}\,dx=0 \] is

• A. \(y^{3}=3x^{4}+cx\)
• B. \(y^{3}=x^{3}\log|x|+c\)
• C. \(e^{\left(\frac{y^{3}}{x^{3}}\right)}=cx^{3}\)
• D. \(y^{3}=\log(cx)^{3}\)

Hint:

If the differential equation can be written in terms of \[ \frac{y}{x}, \] use the substitution \[ \boxed{y=vx.} \] This converts the equation into a separable differential equation.

Answer 1

The Correct Option is C

Step 1: Rewrite the differential equation. Given, \[ x^{3}\,dx-xy^{2}\,dy+y^{3}\,dx=0. \] Combining the \(dx\)-terms, \[ (x^{3}+y^{3})\,dx = xy^{2}\,dy. \] Hence, \[ \frac{dy}{dx} = \frac{x^{3}+y^{3}}{xy^{2}} = \frac{x^{2}}{y^{2}}+\frac{y}{x}. \]

Step 2: Reduce to a homogeneous equation. Let \[ v=\frac{y}{x}, \] so that \[ y=vx, \qquad \frac{dy}{dx} = v+x\frac{dv}{dx}. \] Substituting, \[ v+x\frac{dv}{dx} = \frac1{v^{2}}+v. \] Thus, \[ x\frac{dv}{dx} = \frac1{v^{2}}. \] Separating variables, \[ v^{2}\,dv = \frac{dx}{x}. \]

Step 3: Integrate both sides. Integrating, \[ \frac{v^{3}}3 = \log|x|+C. \] Therefore, \[ v^{3} = 3\log|x|+C_1. \] Since \[ v=\frac{y}{x}, \] we obtain \[ \frac{y^{3}}{x^{3}} = 3\log|x|+C_1. \] Exponentiating, \[ e^{\frac{y^{3}}{x^{3}}} = e^{C_1}e^{3\log|x|} = Cx^{3}, \] where \(C=e^{C_1}\). Hence, \[ \boxed{ e^{\left(\frac{y^{3}}{x^{3}}\right)} = Cx^{3}. } \] Therefore, the correct option is \(\boxed{(C)}\).