Question

Solve the differential equation $y \, dx + (x - y^3) \, dy = 0$.

Hint:

If a differential equation looks complicated and has an expression like $(x \pm y^n)dy$, it is almost always a linear differential equation of the form $\frac{dx}{dy} + Px = Q$. Inverting the derivative from $\frac{dy}{dx}$ to $\frac{dx}{dy}$ simplifies the algebra instantly!

Answer 1

Concept: The given differential equation can be tested for standard structures (separable variables, homogeneous, or linear forms). Let us rearrange the expression to see its mathematical form: \[ y \, dx + (x - y^3) \, dy = 0 \] If we try to isolate $\frac{dy}{dx}$, we get: \[ (x - y^3) \, dy = -y \, dx \implies \frac{dy}{dx} = \frac{-y}{x - y^3} \] This form is non-separable and non-linear in terms of $y$ because of the $y^3$ term in the denominator. Let us instead invert the derivative to find $\frac{dx}{dy}$: \[ y \, dx = (y^3 - x) \, dy \implies \frac{dx}{dy} = \frac{y^3 - x}{y} \] Splitting the fraction on the right-hand side gives: \[ \frac{dx}{dy} = \frac{y^3}{y} - \frac{x}{y} \implies \frac{dx}{dy} = y^2 - \frac{x}{y} \] Rearranging terms yields: \[ \frac{dx}{dy} + \frac{1}{y} \cdot x = y^2 \] This matches the standard mathematical structure of a First-Order Linear Differential Equation in $x$, which is expressed generally as: \[ \frac{dx}{dy} + P(y)x = Q(y) \] Here, by matching components, we find: \[ P(y) = \frac{1}{y} \quad \text{and} \quad Q(y) = y^2 \]

Step 1: Calculating the Integrating Factor (I.F.)
The integrating factor for a linear differential equation of this type is computed using the formula: \[ \text{I.F.} = e^{\int P(y) \, dy} \] Substituting our value $P(y) = \frac{1}{y}$: \[ \text{I.F.} = e^{\int \frac{1}{y} \, dy} \] Since the antiderivative $\int \frac{1}{y} \, dy = \ln|y|$, we can substitute this back: \[ \text{I.F.} = e^{\ln|y|} = y \]

Step 2: Formulating the general solution equation
The standard general solution formula for a linear differential equation in $x$ is given by: \[ x \times (\text{I.F.}) = \int Q(y) \times (\text{I.F.}) \, dy + C \] Substituting our expressions for $\text{I.F.} = y$ and $Q(y) = y^2$ into this equation, we obtain: \[ x \cdot y = \int (y^2 \cdot y) \, dy + C \] Simplifying the integrand: \[ xy = \int y^3 \, dy + C \]

Step 3: Performing the integration
Using the basic power rule of integration $\int y^n \, dy = \frac{y^{n+1}}{n+1}$ for $n=3$: \[ xy = \frac{y^4}{4} + C \]

Step 4: Expressing $x$ explicitly
To completely isolate $x$ and match the standard choices, we divide both sides of the solution equation by $y$ (assuming $y \ne 0$): \[ x = \frac{\frac{y^4}{4} + C}{y} \] \[ x = \frac{y^3}{4} + \frac{C}{y} \]