Consider the fourth-order, linear differential operator et D+ D² + 9 – t 16 + t4 2t D3 4+t² sin(2t) L(t) = Dª – 36 – t2 sin(t) | (1) Evaluate L(t)et. (2) Give the interval of definition for the solution of the initial-value problem e2t L(t)y = y" (7) = y"(7) = y (7) = y(7) = 0 . %3D %3D 8 +t (3) Give the interval of definition for the solution of the initial-value problem e2t L(t)y : y" (-7) = y" (-7) = y'(-7) = y(-7) = 5. 8 +t'

Activity 3: The car body vertical displacement, y(t), can be approximately modelled as a damped mass-spring system (called car suspension system) using the following differential equation: Car Body Attaches Here Spring and Damper Hub Assembly ННИНЯТ d²y + b₂ dt² Wheel and Tire dy dt + ky = f(t) k Car body mcar 4 bv y (t) f(t) 7/1 Where y(t) is the vertical displacement, mear is the car mas, b, is the damping coefficient, k is the spring stiffness constant, t is the time, and f(t) is the hight of the street from the ground level as a function of time. Given that m = 1500 (Kg), b₂ = 2000 (N.s/m), and k = 100000 (N/m). Answer the following questions: (3-A) If the street is straight, so the f(t) = 0, determine the general solution of y(t). (3-B) Given the initial conditions y(0) = 2, and y'(0) = 1, determine the particular solution of y(t). (3-C) Use Laplace transform to solve the given differential equation with the same initial conditions given in (3-B).

can you please provide explanations

Solve the differential equation in Exercises 9–22. dy dy = x²Vy, y> 0 10. dx 9. 2Vxy = 1, x, y > 0 dx dy 11. dx dy 3x e 12. dx dy dy 13. Vycos? Vy dx 14. V2ry dx dy = ev+sin x dx dy 15. Vx dx x > 0 16. (sec x)- dy 2rV1 – y², -1< y < 1 17. dx dy 18. dx e2r-y er+y dy 19. y? Зx2уз —бх? dx dy 20. — ху + 3x — 2у — 6 dx 1 dy х dx dy 22. dx ye + 2Vye 21. et-y + e* + e¯y + 1 %3D