Given an IVP an(x)dnydxn+an−1(x)dn−1ydxn−1+…+a1( x)dydx+a0(x)y=g(x)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−1y(x0)=y0, y′(x0)=y1, ⋯, y(n−1)(x0)=yn−1 If the coefficients an(x),…,a0(x)an(x),…,a0(x) and the right hand side of the equation g(x)g(x) are continuous on an interval II and if an(x)≠0an(x)≠0 on II then the IVP has a unique solution for the point x0∈Ix0∈I that exists on the whole interval II. Consider the IVP on the whole real line sin(x)d2ydx2+cos(x)dydx+sin(x)y=tan (x)sin⁡(x)d2ydx2+cos⁡(x)dydx+sin⁡(x )y=tan⁡(x) y(1)=11, y′(1)=4,y(1)=11, y′(1)=4, The Fundamental Existence Theorem for Linear Differential Equations guarantees the existence of a unique solution on the interval