1. The problem statement, all variables and given/known data P4-1. The Method of Frobenius: Sines and Cosines. The solutions to the differential equation y"+ y = 0 can be expressed in terms of our familiar sine and cosine: y(x) = Acos(x) + Bsin(x) . Use the Method of Frobenius to solve the above differential equation for the even solution (cosine) and odd solution (sine). By doing so, express the cosine and sine each as power series. 2. Relevant equations 3. The attempt at a solution I think the first thing I would have to do is convert my trigonometric function to a power series function. sin(x)=x-x^3/3!+x^5/5!+... , cos(x)=1-x^2/2!+x^4/4! +... and y(x) now equals : y(x)=x-x^3/3!+x^5/5!+1-x^2/2!+x^4/4! +... and now I can apply the Method of frobenius? y''+y=A(-1)^n/(2n!)(x^2n+(2n(2n-1)*x^2n-1)+B(-1)^n/(2n+1!)*(x^2n+1 +(2n)(2n+1)x^2n-1) where sin(x)=(-1)^n*x^2n/(2n)! , and cos(x)=(-1)^n*x^2n/(2n+1)!. My expression does not look like it would be equal to zero.