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Solve the second-order ordinary differential equation:

\(y''(t) + 4y'(t) + 4y(t) = 4e^t\)

MaxWong  Nov 12, 2017

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 #2
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My solution in LaTeX:(I guess it is the same...)

\(y'' + 4y' + 4y = 4e^t\\ \text{Let }y_h \text{ be the homogeneous case.}\\ y_h'' + 4y_h' + 4y_h = 0\\ \lambda^2 + 4\lambda + 4 = 0\\ \lambda = -2\\ \therefore y_h = C_1 e^{-2t} + C_2 te^{-2t}\\ \text{Let } y_p = Ae^t.\\ y_p=y'_p =y''_p= Ae^t\\ Ae^t + 4Ae^t + 4Ae^t = 4e^t\\ 9A = 4\\ A = \dfrac{4}{9}\\ \therefore y_p = \dfrac{4e^t}{9}\\ \therefore y = y_h + y_p = C_1 e^{-2t} + C_2 te^{-2t} + \dfrac{4e^t}{9}\)

:)

MaxWong  Nov 12, 2017
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 #1
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Max: Compare your solution to this by "Mathematica 11 Home Edition" !!

 

 

Solve 4 ( dy(t))/( dt) + ( d^2 y(t))/( dt^2) + 4 y(t) = 4 e^t:

The general solution will be the sum of the complementary solution and particular solution.
Find the complementary solution by solving ( d^2 y(t))/( dt^2) + 4 ( dy(t))/( dt) + 4 y(t) = 0:

Assume a solution will be proportional to e^(λ t) for some constant λ.
Substitute y(t) = e^(λ t) into the differential equation:
( d^2 )/( dt^2)(e^(λ t)) + 4 ( d)/( dt)(e^(λ t)) + 4 e^(λ t) = 0

Substitute ( d^2 )/( dt^2)(e^(λ t)) = λ^2 e^(λ t) and ( d)/( dt)(e^(λ t)) = λ e^(λ t):
λ^2 e^(λ t) + 4 λ e^(λ t) + 4 e^(λ t) = 0

Factor out e^(λ t):
(λ^2 + 4 λ + 4) e^(λ t) = 0

Since e^(λ t) !=0 for any finite λ, the zeros must come from the polynomial:
λ^2 + 4 λ + 4 = 0

Factor:
(2 + λ)^2 = 0

Solve for λ:
λ = -2 or λ = -2

The multiplicity of the root λ = -2 is 2 which gives y_1(t) = c_1 e^(-2 t), y_2(t) = c_2 e^(-2 t) t as solutions, where c_1 and c_2 are arbitrary constants.
The general solution is the sum of the above solutions:
y(t) = y_1(t) + y_2(t) = c_1/e^(2 t) + (c_2 t)/e^(2 t)

Determine the particular solution to ( d^2 y(t))/( dt^2) + 4 y(t) + 4 ( dy(t))/( dt) = 4 e^t by the method of undetermined coefficients:
The particular solution to ( d^2 y(t))/( dt^2) + 4 y(t) + 4 ( dy(t))/( dt) = 4 e^t is of the form:
y_p(t) = a_1 e^t

Solve for the unknown constant a_1:
Compute ( dy_p(t))/( dt):
( dy_p(t))/( dt) = ( d)/( dt)(a_1 e^t)
 = e^t a_1

Compute ( d^2 y_p(t))/( dt^2):
( d^2 y_p(t))/( dt^2) = ( d^2 )/( dt^2)(a_1 e^t)
 = e^t a_1

Substitute the particular solution y_p(t) into the differential equation:
( d^2 y_p(t))/( dt^2) + 4 ( dy_p(t))/( dt) + 4 y_p(t) = 4 e^t
a_1 e^t + 4 (a_1 e^t) + 4 (a_1 e^t) = 4 e^t

Simplify:
9 a_1 e^t = 4 e^t

Equate the coefficients of e^t on both sides of the equation:
9 a_1 = 4

Solve the equation:
a_1 = 4/9

Substitute a_1 into y_p(t) = a_1 e^t:
y_p(t) = (4 e^t)/9

The general solution is:
y(t) = y_c(t) + y_p(t) = (4 e^t)/9 + c_1/e^(2 t) + (c_2 t)/e^(2 t)

Guest Nov 12, 2017
 #2
avatar+6881 
+1
Best Answer

My solution in LaTeX:(I guess it is the same...)

\(y'' + 4y' + 4y = 4e^t\\ \text{Let }y_h \text{ be the homogeneous case.}\\ y_h'' + 4y_h' + 4y_h = 0\\ \lambda^2 + 4\lambda + 4 = 0\\ \lambda = -2\\ \therefore y_h = C_1 e^{-2t} + C_2 te^{-2t}\\ \text{Let } y_p = Ae^t.\\ y_p=y'_p =y''_p= Ae^t\\ Ae^t + 4Ae^t + 4Ae^t = 4e^t\\ 9A = 4\\ A = \dfrac{4}{9}\\ \therefore y_p = \dfrac{4e^t}{9}\\ \therefore y = y_h + y_p = C_1 e^{-2t} + C_2 te^{-2t} + \dfrac{4e^t}{9}\)

:)

MaxWong  Nov 12, 2017
 #3
avatar+499 
-1

Haha too much nonsence laugh

ProMagma  Nov 12, 2017

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