To solve the initial value problem with dy/dx = x - 2y and y(0) = 1, the student must use an integrating factor to transform the linear differential equation, integrate to find the general solution, and then apply the initial condition to determine the specific curve that is the correct solution.
The student is given an Initial Value Problem (IVP) with a differential equation dy/dx = x - 2y and an initial condition y(0) = 1. To solve this problem, we can use an integrating factor or recognize that this is a linear differential equation. Since there is no list of curves provided in the question, we cannot explicitly choose the correct curve; however, any viable solution must satisfy both the differential equation and the initial condition when evaluated at x=0.
In general, for a first-order linear differential equation of the form dy/dx + P(x)y = Q(x), an integrating factor μ(x) can be used, where μ(x) = e∫ P(x)dx. By multiplying both sides of the differential equation by the integrating factor, it transforms into an exact equation, which can be integrated to find the solution y(x).
Applying this to the provided IVP, we can determine the solution that satisfies y(0) = 1. To find the integrating factor, we integrate the coefficient of y, which is -2, giving us μ(x) = e-2x. Multiplying the original equation by this factor and integrating will result in the general solution. Plugging in the initial value, we can find the specific solution to the IVP.
The student must apply the method of integrating factors to the linear differential equation, and check if the derived function satisfies y(0) = 1, to confirm the correct solution for the IVP.