Question

Given an IVP

an(x)dnydxn+an−1(x)dn−1ydxn−1+…+a1(x)dydx+a0(x)y=g(x)

y(x0)=y0, y′(x0)=y1, ⋯, y(n−1)(x0)=yn−1 If the coefficients an(x),…,a0(x) and the right hand side of the equation g(x) are continuous on an interval I and if an(x)≠0 on I then the IVP has a unique solution for the point x0∈I that exists on the whole interval I. Consider the IVP on the whole real line

sin(x)d2ydx2+cos(x)dydx+sin(x)y=tan(x)

y(0.5)=20, y′(0.5)=3, The Fundamental Existence Theorem for Linear Differential Equations guarantees the existence of a unique solution on the interval

Answer 1

A unique solution to the linear differential equation with continuous coefficients and IVP exists on an interval where sin(x) is not zero, subject to continuity and initial conditions.

Step-by-step explanation:

The student's question pertains to the uniqueness and existence of the solution to a second-order linear differential equation with variable coefficients. When the coefficients an(x), …, a0(x) and the non-homogeneous term g(x) are continuous on an interval I, and an(x) is non-zero on I, the Initial Value Problem (IVP) guarantees a unique solution at a point x0 within I. Given that sin(x), cos(x), and tan(x) are continuous except at isolated points where tan(x) is undefined, a unique solution exists and is continuous within an interval where sin(x) is not zero.

The solution to the differential equation is subject to the principle of superposition if the equation is linear and homogeneous. This principle asserts that a linear combination of solutions to a linear homogeneous differential equation is also a solution, allowing for the construction of more complex solutions from simpler ones.

Answer 2

Step-by-step explanation: see attachment below