Let y = y₁(x) and y = y₂(x) be two distinct solutions of the differential equation dy dx + 2xy = f(x) , where f(x) is a continuous function. If y₁(0) = 2 and y₂(0) = 6 , then the value of x > 0 for which y₂(x) - y₁(x) = 2 is

Step-by-step solution

Let y(x) = y₂(x) - y₁(x) . Since y₁ and y₂ both satisfy the given differential equation, their difference satisfies the homogeneous equation: dy dx + 2xy = 0 Separating the variables, we get: dy y = -2x , dx Integrating both sides: |y| = -x^2 + C₁ y(x) = C e^ -x^2 Using the given initial conditions, y(0) = y₂(0) - y₁(0) = 6 - 2 = 4 . Thus, C = 4 , and the difference function is y(x) = 4e^ -x^2 . We are required to find x > 0 such that y₂(x) - y₁(x) = 2 : 4e^ -x^2 = 2 e^ -x^2 = 1 2 e^ x^2 = 2 Taking the natural loga