Question

Let Q1=y(1.1), Q2=y(1.2), Q3=y(1.3), where y=y(x) solves...

1) y'''+2y''-5y'- 6y=4x^2 where y(0)=1, y'(0)=2, y''(0)=3

2) y'''- 6y''+11y'- 6y=6e^(4x) where y(0)=4, y'(0)=10, y''(0)=30

3) y''- 6y'+9y=4e^(3x) ln(x) where y(1)=, y'(1)=2

Please show all steps and thank you!!!

Answer 1

(1):

Find the complementary solution by solving ( d^3 y(x))/( dx^3) + 2 ( d^2 y(x))/( dx^2) - 5 ( dy(x))/( dx) - 6 y(x) = 0:
Assume a solution will be proportional to e^(λ x) for some constant λ.
Substitute y(x) = e^(λ x) into the differential equation:
( d^3 )/( dx^3)(e^(λ x)) + 2 ( d^2 )/( dx^2)(e^(λ x)) - 5 d/( dx)(e^(λ x)) - 6 e^(λ x) = 0
Substitute ( d^3 )/( dx^3)(e^(λ x)) = λ^3 e^(λ x), ( d^2 )/( dx^2)(e^(λ x)) = λ^2 e^(λ x), and d/( dx)(e^(λ x)) = λ e^(λ x):
λ^3 e^(λ x) + 2 λ^2 e^(λ x) - 5 λ e^(λ x) - 6 e^(λ x) = 0
Factor out e^(λ x):
(λ^3 + 2 λ^2 - 5 λ - 6) e^(λ x) = 0
Since e^(λ x) !=0 for any finite λ, the zeros must come from the polynomial:
λ^3 + 2 λ^2 - 5 λ - 6 = 0
Factor:
(λ - 2) (λ + 1) (λ + 3) = 0
Solve for λ:
λ = -3 or λ = -1 or λ = 2
The root λ = -3 gives y_1(x) = c_1 e^(-3 x) as a solution, where c_1 is an arbitrary constant.
The root λ = -1 gives y_2(x) = c_2 e^(-x) as a solution, where c_2 is an arbitrary constant.
The root λ = 2 gives y_3(x) = c_3 e^(2 x) as a solution, where c_3 is an arbitrary constant.
The general solution is the sum of the above solutions:
y(x) = y_1(x) + y_2(x) + y_3(x) = c_1 e^(-3 x) + c_2 e^(-x) + c_3 e^(2 x)
Determine the particular solution to ( d^3 y(x))/( dx^3) + 2 ( d^2 y(x))/( dx^2) - 5 ( dy(x))/( dx) - 6 y(x) = 4 x^2 by the method of undetermined coefficients:
The particular solution to ( d^3 y(x))/( dx^3) + 2 ( d^2 y(x))/( dx^2) - 5 ( dy(x))/( dx) - 6 y(x) = 4 x^2 is of the form:
y_p(x) = a_1 + a_2 x + a_3 x^2
Solve for the unknown constants a_1, a_2, and a_3:
Compute ( dy_p(x))/( dx):
( dy_p(x))/( dx) = d/( dx)(a_1 + a_2 x + a_3 x^2)
= a_2 + 2 a_3 x
Compute ( d^2 y_p(x))/( dx^2):
( d^2 y_p(x))/( dx^2) = ( d^2 )/( dx^2)(a_1 + a_2 x + a_3 x^2)
= 2 a_3
Compute ( d^3 y_p(x))/( dx^3):
( d^3 y_p(x))/( dx^3) = ( d^3 )/( dx^3)(a_1 + a_2 x + a_3 x^2)
= 0
Substitute the particular solution y_p(x) into the differential equation:
( d^3 y_p(x))/( dx^3) + 2 ( d^2 y_p(x))/( dx^2) - 5 ( dy_p(x))/( dx) - 6 y_p(x) = 4 x^2
2 (2 a_3) - 5 (a_2 + 2 a_3 x) - 6 (a_1 + a_2 x + a_3 x^2) = 4 x^2
Simplify:
-6 a_1 - 5 a_2 + 4 a_3 + (-6 a_2 - 10 a_3) x - 6 a_3 x^2 = 4 x^2
Equate the coefficients of 1 on both sides of the equation:
-6 a_1 - 5 a_2 + 4 a_3 = 0
Equate the coefficients of x on both sides of the equation:
-6 a_2 - 10 a_3 = 0
Equate the coefficients of x^2 on both sides of the equation:
-6 a_3 = 4
Solve the system:
a_1 = -37/27
a_2 = 10/9
a_3 = -2/3
Substitute a_1, a_2, and a_3 into y_p(x) = a_1 + x a_2 + x^2 a_3:
y_p(x) = -(2 x^2)/3 + (10 x)/9 - 37/27
The general solution is:
y(x) = y_c(x) + y_p(x) = -(2 x^2)/3 + (10 x)/9 + c_1 e^(-3 x) + c_2 e^(-x) + c_3 e^(2 x) - 37/27
Solve for the unknown constants using the initial conditions:
Compute ( dy(x))/( dx):
( dy(x))/( dx) = d/( dx)(-(2 x^2)/3 + (10 x)/9 + c_1 e^(-3 x) + c_2 e^(-x) + c_3 e^(2 x) - 37/27)
= -(4 x)/3 - 3 c_1 e^(-3 x) - c_2 e^(-x) + 2 c_3 e^(2 x) + 10/9
Compute ( d^2 y(x))/( dx^2):
( d^2 y(x))/( dx^2) = ( d^2 )/( dx^2)(-(2 x^2)/3 + (10 x)/9 + c_1 e^(-3 x) + c_2 e^(-x) + c_3 e^(2 x) - 37/27)
= 9 c_1 e^(-3 x) + c_2 e^(-x) + 4 c_3 e^(2 x) - 4/3
Substitute y(0) = 1 into y(x) = -(2 x^2)/3 + (10 x)/9 + e^(-3 x) c_1 + e^(-x) c_2 + e^(2 x) c_3 - 37/27:
c_1 + c_2 + c_3 - 37/27 = 1
Substitute y'(0) = 2 into ( dy(x))/( dx) = -(4 x)/3 - 3 e^(-3 x) c_1 - e^(-x) c_2 + 2 e^(2 x) c_3 + 10/9:
-3 c_1 - c_2 + 2 c_3 + 10/9 = 2
Substitute y''(0) = 3 into ( d^2 y(x))/( dx^2) = 9 e^(-3 x) c_1 + e^(-x) c_2 + 4 e^(2 x) c_3 - 4/3:
9 c_1 + c_2 + 4 c_3 - 4/3 = 3
Solve the system:
c_1 = -7/54
c_2 = 3/2
c_3 = 1
Substitute c_1 = -7/54, c_2 = 3/2, and c_3 = 1 into y(x) = -(2 x^2)/3 + (10 x)/9 + e^(-3 x) c_1 + e^(-x) c_2 + e^(2 x) c_3 - 37/27:
Answer: |
| y(x) = 1/54 (54 e^(2 x) + 81 e^(-x) - 7 e^(-3 x) - 36 x^2 + 60 x - 74)

(2):

Find the complementary solution by solving ( d^3 y(x))/( dx^3) - 6 ( d^2 y(x))/( dx^2) + 11 ( dy(x))/( dx) - 6 y(x) = 0:
Assume a solution will be proportional to e^(λ x) for some constant λ.
Substitute y(x) = e^(λ x) into the differential equation:
( d^3 )/( dx^3)(e^(λ x)) - 6 ( d^2 )/( dx^2)(e^(λ x)) + 11 d/( dx)(e^(λ x)) - 6 e^(λ x) = 0
Substitute ( d^3 )/( dx^3)(e^(λ x)) = λ^3 e^(λ x), ( d^2 )/( dx^2)(e^(λ x)) = λ^2 e^(λ x), and d/( dx)(e^(λ x)) = λ e^(λ x):
λ^3 e^(λ x) - 6 λ^2 e^(λ x) + 11 λ e^(λ x) - 6 e^(λ x) = 0
Factor out e^(λ x):
(λ^3 - 6 λ^2 + 11 λ - 6) e^(λ x) = 0
Since e^(λ x) !=0 for any finite λ, the zeros must come from the polynomial:
λ^3 - 6 λ^2 + 11 λ - 6 = 0
Factor:
(λ - 3) (λ - 2) (λ - 1) = 0
Solve for λ:
λ = 1 or λ = 2 or λ = 3
The root λ = 1 gives y_1(x) = c_1 e^x as a solution, where c_1 is an arbitrary constant.
The root λ = 2 gives y_2(x) = c_2 e^(2 x) as a solution, where c_2 is an arbitrary constant.
The root λ = 3 gives y_3(x) = c_3 e^(3 x) as a solution, where c_3 is an arbitrary constant.
The general solution is the sum of the above solutions:
y(x) = y_1(x) + y_2(x) + y_3(x) = c_1 e^x + c_2 e^(2 x) + c_3 e^(3 x)
Determine the particular solution to ( d^3 y(x))/( dx^3) - 6 ( d^2 y(x))/( dx^2) + 11 ( dy(x))/( dx) - 6 y(x) = 6 e^(4 x) by the method of undetermined coefficients:
The particular solution to ( d^3 y(x))/( dx^3) - 6 ( d^2 y(x))/( dx^2) + 11 ( dy(x))/( dx) - 6 y(x) = 6 e^(4 x) is of the form:
y_p(x) = a_1 e^(4 x)
Solve for the unknown constant a_1:
Compute ( dy_p(x))/( dx):
( dy_p(x))/( dx) = d/( dx)(a_1 e^(4 x))
= 4 a_1 e^(4 x)
Compute ( d^2 y_p(x))/( dx^2):
( d^2 y_p(x))/( dx^2) = ( d^2 )/( dx^2)(a_1 e^(4 x))
= 16 a_1 e^(4 x)
Compute ( d^3 y_p(x))/( dx^3):
( d^3 y_p(x))/( dx^3) = ( d^3 )/( dx^3)(a_1 e^(4 x))
= 64 a_1 e^(4 x)
Substitute the particular solution y_p(x) into the differential equation:
( d^3 y_p(x))/( dx^3) - 6 ( d^2 y_p(x))/( dx^2) + 11 ( dy_p(x))/( dx) - 6 y_p(x) = 6 e^(4 x)
64 a_1 e^(4 x) - 6 (16 a_1 e^(4 x)) + 11 (4 a_1 e^(4 x)) - 6 (a_1 e^(4 x)) = 6 e^(4 x)
Simplify:
6 a_1 e^(4 x) = 6 e^(4 x)
Equate the coefficients of e^(4 x) on both sides of the equation:
6 a_1 = 6
Solve the equation:
a_1 = 1
Substitute a_1 into y_p(x) = a_1 e^(4 x):
y_p(x) = e^(4 x)
The general solution is:
y(x) = y_c(x) + y_p(x) = e^(4 x) + c_1 e^x + c_2 e^(2 x) + c_3 e^(3 x)
Solve for the unknown constants using the initial conditions:
Compute ( dy(x))/( dx):
( dy(x))/( dx) = d/( dx)(e^(4 x) + c_1 e^x + c_2 e^(2 x) + c_3 e^(3 x))
= 4 e^(4 x) + c_1 e^x + 2 c_2 e^(2 x) + 3 c_3 e^(3 x)
Compute ( d^2 y(x))/( dx^2):
( d^2 y(x))/( dx^2) = ( d^2 )/( dx^2)(e^(4 x) + c_1 e^x + c_2 e^(2 x) + c_3 e^(3 x))
= 16 e^(4 x) + c_1 e^x + 4 c_2 e^(2 x) + 9 c_3 e^(3 x)
Substitute y(0) = 4 into y(x) = e^(4 x) + e^x c_1 + e^(2 x) c_2 + e^(3 x) c_3:
c_1 + c_2 + c_3 + 1 = 4
Substitute y'(0) = 10 into ( dy(x))/( dx) = 4 e^(4 x) + e^x c_1 + 2 e^(2 x) c_2 + 3 e^(3 x) c_3:
c_1 + 2 c_2 + 3 c_3 + 4 = 10
Substitute y''(0) = 30 into ( d^2 y(x))/( dx^2) = 16 e^(4 x) + e^x c_1 + 4 e^(2 x) c_2 + 9 e^(3 x) c_3:
c_1 + 4 c_2 + 9 c_3 + 16 = 30
Solve the system:
c_1 = 1
c_2 = 1
c_3 = 1
Substitute c_1 = 1, c_2 = 1, and c_3 = 1 into y(x) = e^(4 x) + e^x c_1 + e^(2 x) c_2 + e^(3 x) c_3:
Answer: |
| y(x) = e^x (e^(3 x) + e^(2 x) + e^x + 1)

(3):

Find the complementary solution by solving ( d^2 y(x))/( dx^2) - 6 ( dy(x))/( dx) + 9 y(x) = 0:
Assume a solution will be proportional to e^(λ x) for some constant λ.
Substitute y(x) = e^(λ x) into the differential equation:
( d^2 )/( dx^2)(e^(λ x)) - 6 d/( dx)(e^(λ x)) + 9 e^(λ x) = 0
Substitute ( d^2 )/( dx^2)(e^(λ x)) = λ^2 e^(λ x) and d/( dx)(e^(λ x)) = λ e^(λ x):
λ^2 e^(λ x) - 6 λ e^(λ x) + 9 e^(λ x) = 0
Factor out e^(λ x):
(λ^2 - 6 λ + 9) e^(λ x) = 0
Since e^(λ x) !=0 for any finite λ, the zeros must come from the polynomial:
λ^2 - 6 λ + 9 = 0
Factor:
(λ - 3)^2 = 0
Solve for λ:
λ = 3 or λ = 3
The multiplicity of the root λ = 3 is 2 which gives y_1(x) = c_1 e^(3 x), y_2(x) = c_2 e^(3 x) x as solutions, where c_1 and c_2 are arbitrary constants.
The general solution is the sum of the above solutions:
y(x) = y_1(x) + y_2(x) = c_1 e^(3 x) + c_2 e^(3 x) x
Determine the particular solution to ( d^2 y(x))/( dx^2) - 6 ( dy(x))/( dx) + 9 y(x) = 4 e^(3 x) log(x) by variation of parameters:
List the basis solutions in y_c(x):
y_(b_1)(x) = e^(3 x) and y_(b_2)(x) = e^(3 x) x
Compute the Wronskian of y_(b_1)(x) and y_(b_2)(x):
W(x) = left bracketing bar e^(3 x) | e^(3 x) x
d/( dx)(e^(3 x)) | d/( dx)(e^(3 x) x) right bracketing bar = left bracketing bar e^(3 x) | e^(3 x) x
3 e^(3 x) | e^(3 x) + 3 e^(3 x) x right bracketing bar = e^(6 x)
Let f(x) = 4 e^(3 x) log(x):
Let v_1(x) = - integral(f(x) y_(b_2)(x))/W(x) dx and v_2(x) = integral(f(x) y_(b_1)(x))/W(x) dx:
The particular solution will be given by:
y_p(x) = v_1(x) y_(b_1)(x) + v_2(x) y_(b_2)(x)
Compute v_1(x):
v_1(x) = - integral4 x log(x) dx = -4 (-x^2/4 + 1/2 x^2 log(x))
Compute v_2(x):
v_2(x) = integral4 log(x) dx = 4 (-x + x log(x))
The particular solution is thus:
y_p(x) = v_1(x) y_(b_1)(x) + v_2(x) y_(b_2)(x) = -4 e^(3 x) (-x^2/4 + 1/2 x^2 log(x)) + 4 e^(3 x) x (-x + x log(x))
Simplify:
y_p(x) = e^(3 x) x^2 (2 log(x) - 3)
The general solution is given by:
Answer: |
| y(x) = y_c(x) + y_p(x) = c_1 e^(3 x) + c_2 e^(3 x) x + e^(3 x) x^2 (2 log(x) - 3)