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avatar+9575 
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Nov 12, 2017
 #3
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Nov 12, 2017
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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)

Nov 12, 2017
 #2
avatar+536 
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Nov 12, 2017
 #2
avatar+536 
0
Nov 12, 2017
 #2
avatar+118608 
+1

I have never done anything like this before so it was a little difficult for me too.

There are some typo style errors in the online derivation which certainly would have helped confuse you.   surprise

 

Each term if the product of an AP term and a GP term

The AP is             a,       a+d,             a+2d,     ….       a+(n-1)d     ….

The GP is             b,         br,             br^2,     ……      br^(n-1), …..

The AP-GP is     ab,   (a+d)br,    (a+2d)br^2,   ……  [a+(n-1)d]br^n-1 ….

 

Now you want the sum to n terms.

 

 

\(S_n=ab+(a+d)br \;+(a+2d)br^2 +(a+3d)br^3\;+\; \dots\;+[a+(n-1)d]br^{n-1}\\ \text{multiply both sides by r}\\ rS_n=\qquad \qquad abr \;+\; (a+d)br^2 \;+\; (a+2d)br^3 \;+\; \dots\;\;+[a+(n-2)d]br^{n-1}+\;\; [a+(n-1)d]br^{n}\\ subtract\\ (1-r)S_n=ab+br(a+d-a)+br^2(a+2d-a-d)+...+br^{n-1}[a+(n-1)d-\{a+(n-2)d\}]-[a+(n-1)d]br^{n}\\ (1-r)S_n=ab+br(d)+br^2(2d-d)+...+br^{n-1}[(n-1)d-\{(n-2)d\}]-[a+(n-1)d]br^{n}\\ (1-r)S_n=ab+br(d)+br^2(2d-d)+...+br^{n-1}[(nd-d-nd+2d]-[a+(n-1)d]br^{n}\\ (1-r)S_n=ab+brd+br^2d+...+br^{n-1}d-[a+(n-1)d]br^{n}\\ (1-r)S_n=ab+bdr+bdr^2+...+bdr^{n-1}-[a+(n-1)d]br^{n}\\~\\ (1-r)S_n=ab+[bdr+bdr^2+...+bdr^{n-1}]-[a+(n-1)d]br^{n}\\ \\~\\(1-r)S_n=ab+[sum\;of\;GP]\qquad \qquad -[a+(n-1)d]br^{n}\\ \)

 

 

Let's look at the sum of a GP part

\(bdr+bdr^2+...+bdr^{n-1}\\ \text{The first term is bdr and the common ratio is r so}\\ =(bdr)+(bdr)r+...+(bdr)r^{n-2}\\ \text{So the last term is the n-1 term}\\ \text{so in the formula for the sum of a GP r is r, a is replaced with bdr, and n is replaced with n-1}\\ =\frac{bdr(1-r^{n-1})}{1-r}\\ =\frac{bdr(\frac{r-r^n}{r})}{1-r}\\ =\frac{bd(r-r^n)}{1-r}\\ =\frac{bdr(1-r^{n-1})}{1-r}\\ so\\ (1-r)S_n=ab+[sum\;of \;a\;GP]-[a+(n-1)d]br^{n}\\ (1-r)S_n=ab+[\frac{bdr(1-r^{n-1})}{1-r}]-[a+(n-1)d]br^{n}\\ \)

 

 

\((1-r)S_n=ab+[\frac{bdr(1-r^{n-1})}{1-r}]-[a+(n-1)d]br^{n}\\~\\ S_n=\dfrac{ab}{1-r}+\dfrac{bdr(1-r^{n-1})}{(1-r)^2}-\dfrac{[a+(n-1)d]br^{n}}{1-r}\\\)

 

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Oh dear, some of my latex has been trucated. surprise sad

I think you should still be able to work it out but let me know if you cannot follow it :/    laugh

Nov 12, 2017
 #4
avatar+536 
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Nov 12, 2017
 #7
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Nov 12, 2017

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