Hello I need help with these calc 2 questions ty

2. [0.25/1 Points] DETAILS PREVIOUS ANSWERS SCALCET8 9.1.002. Verify that y = -3t cos(t) - 3t is a solution of the following initial-value problem. dy = y + 3t2 sin(t) y(x) = 0 y = -3t cos(t) - 3t = dy X dt LHS = t dt = 3t2 sin(t) - 3t cos(t) - 3t X + y = RHS, so y is a solution of the differential equation. Also y() = 0 V , so the initial condition is satisfied. Need Help? Read It4. [-/1 Points] DETAILS SCALCET8 9.1.004. (a) For what values of k does the function y = cos(kt) satisfy the differential equation 64y" = -81y? (Enter your answers as a comma-separated list.) K = (b) For those values of k, verify that every member of the family of functions y = A sin(kt) + B cos(kt) is also a solution. y = A sin(kt) + B cos(kt) => y' = Ak cos(kt) - Bk sin(kt) => y" = -Ak2 sin(kt) - BK2 cos(kt). The given differential equation 64y" = -81y is equivalent to 64y" + 81y = Thus, LHS = 64y" + 81y = 64(-Ak2 sin(kt) - BK2 cos(kt)) + 81 = -64Ak2 sin(kt) - 64BK2 cos(kt) + sin(kt) + 81B cos(kt) = (81 - 64k ) A sin(kt) + cos(kt) since k-=EXAMPLE 2 Find the solution of the differential equation y' = %(Y2 1) that satisfies the initial condition y(0) = 10. SOLUTION Substituting the values t = 0 and y = 10 into the formula 1+cet y: t 1ce from this example, we get 0 1 + ce 1 ce _ 1 + c _ 1 cl Solving this equation for c, we get 10 10c = 1 + c, which gives it = 1 x . So the solution of the initial value problem is 1 + x et Y = t 1 x e x + x e