Tutorial Sheet_03_UMA023.pdf

Department of Mathematics Thapar Institute of Engineering & Technology, Patiala, UMA 023 (Differential Equations & Linear Algebra): Tutorial Sheet 03 1. Find the general solution of the following differential equations by the method of undetermined coefficients....

Department of Mathematics Thapar Institute of Engineering & Technology, Patiala, UMA 023 (Differential Equations & Linear Algebra): Tutorial Sheet 03 1. Find the general solution of the following differential equations by the method of undetermined coefficients. (i) y ′′ −3y ′ −10y = 1+x2 (ii) 4y ′′ −y = ex +e3x (iii) 3y ′′ +2y ′ −y = −2x e +x (iv) y ′′ + 3y ′ + 2y = cos x + sin x (v) y ′′ + 16y = 16 sin 4x (vi) y ′′ + 2y ′ + 10y = −x ′′ ′ −2x 2 ′′ ′ e sin 3x (vii) y + 4y + 4y = 6e cos x (viii) y + y = 10x4 + 2 2. Solve each differential equation by variation of parameters method. (i) y ′′ +y = sec x (ii) y ′′ +y = sin x (iii) y ′′ +y = cos2 x (iv) xy ′′ −4xy = e2x (v) y ′′ +3y ′ +2y = 1+e 1 x (vi) y ′′ +3y ′ +2y = sin ex (vii) y ′′ + 2y ′ + y = e−x ln x (viii) 3y ′′ − 6y ′ + 6y = ex sec x 3. Obtain the particular solution of the following differential equations by using operator methods. (i) y ′′ − 4y = e2x (ii) y ′′ − y = x2 e2x (iii) y ′′ − y ′ + y = x3 − 3x2 + 1 (iv) y ′′′ − y ′′ = 12x − 2 (v) y ′′ + 2y ′ + y = 2x2 e−2x + 3e2x 4. Find the general solution of the following Cauchy-Euler differential equations. (i) x2 y ′′ + xy ′ − y = 24x (ii) x2 y ′′ − xy ′ − y = ln x 5. An e.m.f. of 3 sin t is applied in LCR circuit. Find the charge(q) and the current(i) in the circuit if q(0) = 0, i(0) = 0, L = 1H, R = 1Ω and C = 0.25F. Answers: 1. (i) y = Ae−2x + Be5x − (50x2 − 30x + 69)/500 (ii) y = Aex/2 + Be−x/2 + (35ex + 3e3x )/105 −x −2x x/3 (iii) y = Ae + Be + (e − 7x − 14)/7 (iv) y = Ae−x + Be−2x + (2 sin x − cos x)/5 (v) y = A cos 4x + B sin 4x − 2x cos x (vi) y = e (A cos 3x + B sin 3x) − (xe−x cos 3x)/6 −x (vii) y = (Ax + B)e−2x + 3e−2x (2x2 − cos 2x)/4 (viii) y = A + Be−x + 2x5 − 10x4 + 40x3 − 120x2 + 242x 2. (i) y = c1 cos x + c2 sin x + x sin x + cos x ln | cos x| (ii) y = c1 cos x + c2 sin x − 21 x cos x Rx e4t   −2x 1 −2x (iii) y = c1 cos x+c2 sin x+ 12 − 16 cos 2x 2x (iv) y = c1 e +c2 e 2x + 4 e ln |x|−e dt , x0 > x0 t 0 (v) y = c1 e−x + c2 e−2x + (e−x + e−2x ) ln(1 + ex ) (vi) y = c1 e−2x + c2 e−x − e−2x sin ex (vii) y = c1 e−x + c2 xe−x + 21 x2 e−x ln x − 34 x2 e−x (viii) y = c1 ex sin x + c2 ex cos x + 31 xex sin x + 1 x 3 e cos x ln | cos x|   x 1 1
3. (i) yp (x) = − e2x (ii) yp (x) = 9x2 − 24x + 26 e2x (iii) yp (x) = x3 − 6x − 5 4 16 27 (iv) yp (x) = −2x3 − 5x2 − 10x − 10 (v) yp (x) = 2e−2x (x2 + 4x + 6) + (1/3)e2x 4. (i) y(x) = c1 x + c2 x−1 + 12x ln x (ii) y(x) = c1 cos 4 ln x + c2 sin 4 ln x − (1/16) ln x