Just (f) please

Recall from calculus that a function can be defined by a power series, for example: exp(t)=n=0n!1tn,cos(t)=n=0(2n)!(1)nt2n If we have a function that is defined by a power series, f(t)=n=0antn where an are the coefficients in R, then we can quickly calculate the power series of it's derivatives: f(t)=n=0(n+1)an+1tn,f(t)=n=0(n+1)(n+2)an+2tn,etc. (a) Verify that (t)=et is a solution of the ODEy=y using power series. (hint: find the power series for each side and verify they are equal by checking that the coefficients on the same powers of t are equal) (b) Suppose f(t)=n=0antn is a solution to the ODE y=3y. Find a relationship between the coefficients an and an+1. (c) Suppose further that f(t) is a solution for the initial value problem y(0)=1. With this information, what are the coefficients an of f ? (d) We know that the solution of y=3y,y(0)=1 is the function e3t. Explain how this gives the same answer as part c. (e) Suppose that (t)=n=0bntn is the solution of the ODE y+(3t2+1)y=cost,y(0)=0. Find bn for n=0,1,2,3,4. (f) Solve the ODE y+(3t2+1)y=cost,y(0)=0 via the method of integrating factors, and graph it on Desmos (hint: Desmos is easily able to anti-differentiate via the fundamental theorem of calculus. See https://www desmos.com/calculator/1tpmvn1uct for an example). Also graph the fourth order estimation to (t) given by n=04bntn, and compare