Step 1: Rearranging the equation.
The given differential equation is:
\[ y \, dx + (x - y^2) \, dy = 0 \] We can rewrite this equation as: \[ y \, dx = -(x - y^2) \, dy \] Now, divide both sides by \( y(x - y^2) \) to separate the variables: \[ \frac{dx}{x - y^2} = -\frac{dy}{y} \] Step 2: Integration of both sides.
Now, we integrate both sides of the equation:
On the left-hand side, the integral is with respect to \(x\), and on the right-hand side, it is with respect to \(y\). Let's integrate: For the left-hand side: \[ \int \frac{dx}{x - y^2} = \ln|x - y^2| \] For the right-hand side: \[ \int -\frac{dy}{y} = -\ln|y| \] Step 3: Combining the results.
After integrating, we have: \[ \ln|x - y^2| = -\ln|y| + C \] where \(C\) is the constant of integration.
Exponentiating both sides to eliminate the logarithms, we get: \[ |x - y^2| = \frac{C}{|y|} \] Thus, the general solution to the differential equation is: \[ x - y^2 = \frac{C}{y} \] Final Answer: The general solution of the differential equation is \( x - y^2 = \frac{C}{y} \).
Let \( y = f(x) \) be the solution of the differential equation\[\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}, \quad -1 < x < 1\] such that \( f(0) = 0 \). If \[6 \int_{-1/2}^{1/2} f(x)dx = 2\pi - \alpha\] then \( \alpha^2 \) is equal to ______.
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: