dy%2Fdx%2B2xy%3Dx%E2%88%9A1-x%5E2.jpeg "(1-x^2)dy/dx+2xy=x√1-x^2")
Solving the Differential Equation: (1-x^2)dy/dx + 2xy = x√(1-x^2)
In this article, we will explore the solution to the differential equation:
$(1-x^2)\frac{dy}{dx} + 2xy = x\sqrt{1-x^2}$
Understanding the Equation
Before we dive into solving the equation, let's take a closer look at what's happening here. We have a first-order differential equation, where the derivative of y with respect to x is multiplied by a function of x. The left-hand side consists of two terms: the derivative of y multiplied by (1-x^2), and the product of 2xy. The right-hand side is the given function x√(1-x^2).
Separation of Variables
To solve this differential equation, we can use the method of separation of variables. The first step is to isolate the derivative term by dividing both sides of the equation by (1-x^2):
$\frac{dy}{dx} + \frac{2xy}{1-x^2} = \frac{x\sqrt{1-x^2}}{1-x^2}$
Now, we can move all the x terms to the right-hand side and integrate both sides with respect to x:
$\int \frac{dy}{y} = \int \frac{x\sqrt{1-x^2} - 2x}{1-x^2} dx$
Integration
To evaluate the integral on the right-hand side, we can use substitution. Let t = 1-x^2, which implies dt = -2x dx. Then, we can rewrite the integral as:
$\int \frac{x\sqrt{1-x^2} - 2x}{1-x^2} dx = \int \frac{\sqrt{t} + 1}{t} dt$
Now, we can integrate:
$\int \frac{\sqrt{t} + 1}{t} dt = \frac{2}{3}t^{3/2} + ln|t| + C$
Substituting back t = 1-x^2, we get:
$\int \frac{x\sqrt{1-x^2} - 2x}{1-x^2} dx = \frac{2}{3}(1-x^2)^{3/2} + ln|1-x^2| + C$
Solving for y
Now, we can equate the integral on the left-hand side:
$ln|y| = \frac{2}{3}(1-x^2)^{3/2} + ln|1-x^2| + C$
Exponentiating both sides, we get:
$y = |1-x^2|^{2/3}e^C$
which is the general solution to the differential equation.
Conclusion
In this article, we solved the differential equation (1-x^2)dy/dx + 2xy = x√(1-x^2) using the method of separation of variables. We integrated both sides of the equation and solved for y, arriving at the general solution y = |1-x^2|^{2/3}e^C. This solution is valid for all x in the domain of the differential equation.
