Question

ydx + (x-y^2)dy=0

Answer 1

xy = \frac{y^3}{3} + C

Answer 2

The given differential equation $ydx + (x-y^2)dy=0$ is a first-order differential equation. It can be transformed into a linear differential equation in $x$ with $y$ as the independent variable.

1. Rearrange the equation to isolate $\frac{dx}{dy}$: $ydx = -(x-y^2)dy$ Divide by $dy$: $y\frac{dx}{dy} = -(x-y^2)$ $\frac{dx}{dy} = -\frac{x}{y} + \frac{y^2}{y}$ $\frac{dx}{dy} = -\frac{x}{y} + y$

2. Rewrite in the standard linear form $\frac{dx}{dy} + P(y)x = Q(y)$: $\frac{dx}{dy} + \frac{1}{y}x = y$ Here, $P(y) = \frac{1}{y}$ and $Q(y) = y$.

3. Calculate the integrating factor (IF): $IF = e^{\int P(y) dy} = e^{\int \frac{1}{y} dy} = e^{\log|y|} = |y|$. Assuming $y > 0$, the integrating factor is $y$.

4. Apply the general solution formula for linear equations: $x \cdot (\text{IF}) = \int Q(y) \cdot (\text{IF}) \, dy + C$. $x \cdot y = \int y \cdot y \, dy + C$

5. Integrate to find the solution: $xy = \int y^2 \, dy + C$ $xy = \frac{y^3}{3} + C$