for each of the following curves write down an integral whic

Solution

x=sin(e^t) d/dx x= d/dx sin(e^t) 1 = 0 x=sin(e^t) 1=cos(e^t) d/dx (e^t) 1=cos(e^t)*(t*e^(t-1)) 1=(t*e^(t-1))cos(e^t) dx/dt = d/dt(sin(e^t)) dx/dt = cos(e^t) *d/dt(e^t) dx/dt = (e^t)*(cos(e^t soo First find the general solution to the corresponding homogeneous ODE: y\" + y = 0 This has characteristic equation r^2 + 1 = 0 <=> r = ± i, so the solutions are cos t and sin t. (In general if there are complex conjugate solutions r = a ± ib then the solutions to the ODE are e^a cos bt and e^a sin bt.) So the complementary solution is y = A cos t + B sin t. Now we need to find a particular solution. The function we\'re after is 3 sin 2t + t cos 2t, so we expect it will be of the form y_p = C cos 2t + D sin 2t + E t cos 2t + F t sin 2t None of these terms are present in the complementary solution, so we\'ll go with this. y_p\' = - 2C sin 2t + 2D cos 2t + E cos 2t - 2E t sin 2t + F sin 2t + 2F t cos 2t = (F-2C) sin 2t + (2D+E) cos 2t - 2E t sin 2t + 2F t cos 2t y_p\" = (2F-4C) cos 2t + (-4D-2E) sin 2t - 2E sin 2t - 4E t cos 2t + 2F cos 2t - 4F t sin 2t = (4F-4C) cos 2t - (4D+4E) sin 2t - 4E t cos 2t - 4F t sin 2t So y_p\" + y_p = [(4F-4C) cos 2t - (4D+4E) sin 2t - 4E t cos 2t - 4F t sin 2t] + [C cos 2t + D sin 2t + E t cos 2t + F t sin 2t] = (4F-3C) cos 2t - (3D+4E) sin 2t- 3E t cos t - 3F t sin 2t which must identically equal 3 sin 2t + t cos 2t. Equating coefficients gives us 4F - 3C = 0 3D + 4E = -3 3E = -1 3F = 0 Hence F = C = 0, E = -1/3 and 3D - 4/3 = -3 => D = -5/9. So y_p = (-5/9) sin 2t - (1/3) t cos 2t. So the general solution is the sum of the complementary and particular solutions, i.e. y = A cos t + B sin t - (5/9) sin 2t - (1/3) t cos 2t.