E2101 Sample Problems

y' + (2/t)y = (cost)/t^2 y(pi) = 0

y = (sint)/t^2

ty' + (t+1)y = t y(ln2) = 1

y = (t - 1 + 2e^(-t))/t

y' + 2y = te^(-2t) y(1) = 0

y = (t^2-1)e^(-2t)/2

y'' + 2y' -3y = 0

y = c1e^t + c2e^(-3t)

y'' - 2y' + 5y = 0 y(pi/2) = 0 y'(pi/2) = 2

y = -e^(t-pi/2)sin2t

y' = (1-2x)/y y(1) = -2

y = -sqrt(2x - 2x^2 + 4)

ty' - y = t^2e^(-t) t>0

y = -te^(-t) + ct

y'' + 8y' - 9y = 0 y(1) = 1 y'(1) = 0

y = 0.1exp[-9(t-1)] + 0.9exp[t-1]

y'' - 9y' + 9y = 0

y = c1exp[(9+3√5)t/2] + c2exp[(9-3√5)t/2]

2y' + y = 3t^2

y = ce^(-t/2) + 3t^2 -12t + 24

y' - 2y = t^2 * e^(2t)

y = ce^(2t) + t^3e^(2t)/3

y'' + 6y' + 13y = 0

y = e^(-3t)[c1cos2t + c2sin2t]

y'' + y' + 1.25y = 0 y(0) = 3 y'(0) = 1

y = e^(-t/2)[3cost + 2.5sint]

y'' + y' + 1.25y = 0

y = e^(-t/2)[c1cost + c2sint]

xy' = (1-y^2)^0.5

y = sin[ln|x| + c]

ty' + 2y = sint y(pi/2) = 1

y =. t^(-2)[((pi)^2/4) - 1 - tcost + sint]

y''' + 4y' = t y(0) = 0 y'(0) = 0 y''(0) = 1

y(t) = (3/16)(1-cos(2t))+(1/8)t^2

y''' - 3y'' + 2y' = t + e^t y(0) = 1 y'(0) = -0.25 y''(0) = -1.5

y(t) = 1 + 0.75t + 0.25t^2 - te^t

2y''' - 4y'' - 2y' + 4y = 0

y(t) = c1e^(2t) + c2e^t + c3e^(-t)

y''' - y'' -y' + y = 0

y(t) = c1e^t + c2e^(-t) + c3te^t

y^(6) - 3y^(4) + 3y'' - y = 0

y(t) = c1e^t + c2e^(-t) + c3te^t + c4te^(-t) + c5t^2e^t + c6t^2e^(-t)

(y/x + 6x) + (lnx - 2)y' = 0

ylnx + 3x^2 - 2y = c

Find all eigenvalues and eigenvectors. (3 -2) (4 -1)

λ = 1 +- 2i x^(1) = c(1+i) ( 2 ) x^(2) = c(1-i) ( 2 )

Find all eigenvalues and eigenvectors. (1 0 0) (2 1 -2) (3 2 1)

λ = 1, 1+- 2i x^(1) = c(2) (-3) (2) x^(2) = c(0) (1) (i) x^(3) = c(0) (1) (-i)

Find all eigenvalues and eigenvectors. (5 -1) (3 1)

λ = 4,2 x^(1) = c(1) (1) x^(2) = c(1) (3)

y'' + 5y' + 3y = 0 y(0) = 1 y'(0) = 0

y = 1/26 (13 + 5√13)exp[(-5 + √13)t/2] + 1/26 (13 - 5√13)exp[(-5-√13)t/2]

y'' + 4y = 0 y(0) = 0 y'(0) = 1

y = 1/2sin2t

Determine whether the members of the given set of vectors are linearly independent. If they are linearly dependent, find a linear relation among them. x1 = (1,1,0) x2 = (0,1,1) x3 = (1,0,1)

Linearly Independent

If this matrix is an augmented matrix, find the solution to the system. (1 3 -2 1) (3 2 -1 2) (-1 -3 2 3)

No solution exists.

If this matrix is an augmented matrix, find the solution to the system. (-2 1 2 0 1) (1 -2 1 3 1) (0 -2 2 1 1)

X = x4(-6.5) + (1.5) (-4 ) + ( 1 ) (-4.5) + (1.5) ( 1 ) + (0)

y'' + 9y = t^2e^(3t) + 6

Y = 1/162(1 - 6t + 9t^2)e^(3t) + 2/3 + c1sint(t) + c2cos(t)

y'' - y' - 2y = 2e^(-t)

Y(t) = (-2/3)te^(-t)

y'' - 2y' - 3y = -3te^(-t)

Y(t) = (3t^2/8 + 3t/16)e^(-t) + C1e^(-t) + C2e^(3t)

y'' + y' - 2y = 2t y(0) = 0 y'(0) = 1

Y(t) = -0.5e^(-2t) + e^t - t - 0.5

y'' + 9y = 9sec^2(3t) 0 < t < pi/6

Y(t) = -1 + c1cos(3t) + c2sin(3t) + ln|sec3t + tan3t|sin3t

y'' + 4y = 3csc2t 0 < t < pi/2

Y(t) = -1.5tcos(2t) + c1cos(2t) + c2sin(2t) + 0.75ln|sin2t|sin2t

y'' + 4y = t^2 + 3e^t y(0) = 0 y'(0) = 2

Y(t) = 0.7sing(2t) - (19/40)cos(2t) + t^2/4 - 1/8 + (3/5)e^t

4y'' - 4y' + y = 16e^(t/2)

Y(t) = 2t^2e^(t/2)

y''' - y' = 2sint

Y(t) = C1 + C2e^t + C3e^(-t) + cos(t)

y^(4) + y''' = sin2t

Y(t) = C1 + C2t + C3t^2 + C4e^(-t) + (1/32)cos(2t) - (1/16)sin(2t)

y^(4) + 2y'' + y = 3 + cos2t

Y(t) = C1cos(t) + C2sin(t) + C3tcos(t) + C4tsin(t) + (1/9)cos(2t) + 3

y^(4) - y = 3t + cost

Y(t) = C1sin(t) + C2cos(t) + C3e^t + C4e^(-t) - 3t - 0.25tsin(t)

u'' + w0^2u = coswt w^2 != w0^2

Y(t) = [1/(w0^2-w^2)]cos(wt) + c1sin(w0t) + c2cos(w0t)

y'' + 2y' + 5y = 4e^(-t)cos2t y(0) = 1 y'(0) = 0

Y(t) = e^tcos(2t) + 0.5e^(-t)sin(2t) + te^(-t)sin(2t)

e^x + (e^xcoty + 2ycscy)y' = 0

e^xsiny + y^2 = c

dr/dtheta = r^2/theta r(1) = 2

r = 2/(1 - 2ln(theta))

Solve the initial value problem. Describe the behavior of the solution as t -> infinity. x' = (5 -1)x (3 1) x(0) = (2) (-1)

x = (-3/2)(1)e^(4t) + (7/2)(1)e^(2t) (1) (3)

Solve the initial value problem. Describe the behavior of the solution as t -> infinity. x' = (-2 1)x (-5 4) x(0) = (1) (3)

x = 1/2(1)e^(3t) + (1/2)(1)e^(-t) (5) (1)

Find the general solution. x' = (1 1) (4 -2)

x = c1(-1)e^(-3t) + c2(1)e^(2t) (4) (1)

Find the general solution. x' = (1 -2) x (3 -4)

x = c1(1)e^(-t) + c2(2)e^(-2t) (1) (3)

Find the general solution. x' = (2 -1)x (3 -2)

x = c1(1)e^t + c2(1)e^(-t) (1) (3)

Transform the given equation into a system of first order equations. u'' + 0.5u' + 2u = 3sint

x2' + 0.5x2 + 2x1 = 3sint(t) x1' = x2

Determine whether the members of the given set of vectors are linearly independent. If they are linearly dependent, find a linear relation among them. x1 = (2,1,0) x2 = (0,1,0) x3 = (-1,2,0)

x3 = -0.5x1 + 2.5x2

Determine whether the members of the given set of vectors are linearly independent. If they are linearly dependent, find a linear relation among them. x1 = (1,1,0) x2 = (0,1,2) x3 = (-1, 2, 0)

x4 = (-7/6)x1 - 0.5x2 - (1/6)x3

y' = xy^3(z+x^2)^(-1/2) y(0) = 1

y = [3 - 2sqrt(1 + x^2)]^(-1/2)

y'' + 5y' = 0

y = c1 + c2e^(-5t)

6y'' - y' -y = 0

y = c1e^(t/2) + c23^(-t/3)

y'' + 2y' - 8y = 0

y = c1e^2t + c2e^(-4t)

Determine whether the members of the given set of vectors are linearly independent. If they are linearly dependent, find a linear relation among them. x1 = (1,2,2,3) x2 = (-1,0,3,1) x3 = (-2,-1,1,0) x4 = (-3,0,-1,3)

x4 = 2x1 - 3x2 + 4x3

Transform the given equation into a system of first order equations. u^(4) - u = 0

x4' - x1 = 0 x1' = x2 x2' = x3 x3' = x4

(2x + 3) + (2y - 2)y' = 0

x^2 + 3x + y^2 - 2y = c

x/(x^2 + y^2)^(3/2) + y/(x^2 + y^2)^(3/2)(dy/dx) = 0

x^2 + y^2 = c

1 + (x/y - siny)y' = 0

xy + ycosy - siny = c

ty' + 2y = sint t>0

y = (c-tcost + sint)/t^2

y' + (1/t)y = 3cos2t

y = (c/t) + (3cos2t)/4t + (3sin2t)/2

If this matrix is an augmented matrix, find the solution to the system. (1 -1 3 3) (2 2 1 2) (1 -1 0 1)

( 5/6 ) (-1/6) ( 2/3 )

(3x^2y + 2xy + y^3) + (x^2 + y^2)y' = 0

(3x^2y + y^3)e^(3x) = c

2x1 + x2 + 3x3 = 1 -x1 + 3x2 - 2x3 = 2 3x1 - 2x2 + x3 = 3 (a) Rewrite the linear system as a matrix equation Ax = b (b) Factorize the coefficient matrix A as A = LU (c) Solve the systems by using L and U.

(a) (2 1 3)(x1) = (1) (-1 3 -2)(x2) = (2) (3 -2 1)(x3) = (3) (b) U = (2 1 3) (-1/2 1 0) (3/2 -1 1) L = (1 0 0) (-1/2 1 0) (3/2 -1 1) (c) y = ( 1 ) (5/2) ( 4 ) x = (12/7) (4/7 ) ( -1 )

2x1 + 2x2 - 3x3 = 1 -3x1 + x2 - 2x3 = 2 x1 - 2x2 + 2x3 = 1 (a) Rewrite the linear system as a matrix equation Ax = b (b) Factorize the coefficient matrix A as A = LU (c) Solve the systems by using L and U.

(a) (2 2 -3)(x1) = (1) (-3 1 -2)(x2) = (2) (1 -2 2)(x3) = (1) (b) U = (2 2 -3) (0 4 -13/2) (0 0 -11/8) L = (1 0 0) (-1.5 1 0) (0.5 -0.75 1) (c) y = (1) (7/2) (25/8) x = (-1/11) (-31/11) (-25/11)

A(t) = (e^t 2e^(-t) e^(2t)) (2e^t e^(-t) -e^(2t)) (-e^t. 3e^(-t) 2e^(2t) B(t) = (2e^t e^(-t) 3e^(2t)) (-e^t 2e^(-t) e^(2t)) (3e^t. -e^(-t) -e^(2t)) find (a) A + 3B (b) AB (c) dA/dt

(a) (7e^t 5e^(-t) 10e^(2t)) (-e^t 7e^(-t) 2e^(2t)) (8e^t 0 -e^(2t)) (b) Check HW 8 Answer Key 1.3.B (c) (e^t -2e^(-t) 2e^t ) (2e^t -e^(-t) -2e^(2t)) (-e^t -3e^(-t) 4e^(2t))

x = ( 2 ) y = (-1 + i) ( 3i ) ( 2 ) (1 - i) (3 - i) find (a) x^Ty (b) y^Ty (c) (x,y) (d) (y,y)

(a) 4i (b) 12 - 8i (c) 2 + 2i (d) 16

(1+t^2)y' + 4ty = (1+t^2)^-2

(arctant + c)/(1+t^2)^2

Compute the determinant (-2 1 0 1) (1 -2 3 0) (0 -2 1 2) (1 0 2 0)

-18

Compute the determinant (2 -1 3) (2 3 1) (1 0 -1)

-18

Compute the determinant (1 3 -2 1) (4 0 -1 2) (-1 1 1 0) (-2 0 3 1)

-22

y' = x(x^2+1)/4y^3 y(0) = -1/sqrt(2)

-sqrt((x^2 + 1)/2)

For the following linear system: 2x1 + x2 = 1 -2x1 + x2 - 2x3 = 2 x1 - 2x2 + 2x3 = 1 1. Rewrite the linear system as a matrix equation Ax = b 2. Use gauss Elimination to solve Ax = b 3. Use Gauss-Jordan method to find A^(-1) 4. Solve the system by using x = A^(-1)b

1. (-2 1 0)(x1) = (1) (-2 1 -2)(x2) = (2) (-1 -2 2)(x3) = (1) 2. x1 = 4, x2 = -7, x3 = -8.5, 3. (1 1 1) (-1 -2 -2) (-1.5 -2.5 -2) 4. ( 4 ) ( -7 ) (-8.5)

y' = (3x^2-1)/(3+2y)

3y + y^2 -x^3 + x = c y != -3/2

dy/dx = x^2/(1+y^2)

3y + y^3-x^3 = c

y' = x^2/y(1 + x^3)

3y^2-2ln|1+x^3| = c