Finding the complementary function of a second order linear differential equation
$begingroup$
I have a question stating
Find the complementary function of $frac{d^2y}{dx^2} + 4frac{dy}{dx} + 40y = x$
So I solved the auxillary equation, $ lambda^2 + 4lambda +40 = 0$, which gives $lambda = -2 pm 6i$. I believe I am correct in thinking this means the complementary solution is of the form $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, where $x in mathbb{R}$. I would have thought this is the final answer but the answer given is $e^{-2x}(Acos{6x} +Bsin{6x})$. To try to get this I did
$$ begin{align}
Ae^{(-2+6i)x} + Be^{(-2-6i)x} &= Ae^{-2x}(cos{6x} +isin{6x}) +Be^{-2x}(cos{6x}+isin{6x}) \ &= e^{-2x}((A+B)cos{6x} + (A-B)isin{6x})
end{align}$$
How do you get to the given answer?
EDIT
To clarify. What I want to know is how to get from my answer, $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, to the textbook's(and wolfram alpha's) answer, $e^{-2x}(Acos{6x} +Bsin{6x})$ ?
calculus ordinary-differential-equations complex-numbers
asked Feb 26 '18 at 16:52
Ewan MillerEwan Miller
12713
$endgroup$
add a comment |
$begingroup$
I have a question stating
Find the complementary function of $frac{d^2y}{dx^2} + 4frac{dy}{dx} + 40y = x$
So I solved the auxillary equation, $ lambda^2 + 4lambda +40 = 0$, which gives $lambda = -2 pm 6i$. I believe I am correct in thinking this means the complementary solution is of the form $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, where $x in mathbb{R}$. I would have thought this is the final answer but the answer given is $e^{-2x}(Acos{6x} +Bsin{6x})$. To try to get this I did
$$ begin{align}
Ae^{(-2+6i)x} + Be^{(-2-6i)x} &= Ae^{-2x}(cos{6x} +isin{6x}) +Be^{-2x}(cos{6x}+isin{6x}) \ &= e^{-2x}((A+B)cos{6x} + (A-B)isin{6x})
end{align}$$
How do you get to the given answer?
EDIT
To clarify. What I want to know is how to get from my answer, $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, to the textbook's(and wolfram alpha's) answer, $e^{-2x}(Acos{6x} +Bsin{6x})$ ?
calculus ordinary-differential-equations complex-numbers
asked Feb 26 '18 at 16:52
Ewan MillerEwan Miller
12713
$endgroup$
$begingroup$
You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
$endgroup$
– LutzL
Feb 26 '18 at 17:34
add a comment |
$begingroup$
I have a question stating
Find the complementary function of $frac{d^2y}{dx^2} + 4frac{dy}{dx} + 40y = x$
So I solved the auxillary equation, $ lambda^2 + 4lambda +40 = 0$, which gives $lambda = -2 pm 6i$. I believe I am correct in thinking this means the complementary solution is of the form $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, where $x in mathbb{R}$. I would have thought this is the final answer but the answer given is $e^{-2x}(Acos{6x} +Bsin{6x})$. To try to get this I did
$$ begin{align}
Ae^{(-2+6i)x} + Be^{(-2-6i)x} &= Ae^{-2x}(cos{6x} +isin{6x}) +Be^{-2x}(cos{6x}+isin{6x}) \ &= e^{-2x}((A+B)cos{6x} + (A-B)isin{6x})
end{align}$$
How do you get to the given answer?
EDIT
To clarify. What I want to know is how to get from my answer, $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, to the textbook's(and wolfram alpha's) answer, $e^{-2x}(Acos{6x} +Bsin{6x})$ ?
calculus ordinary-differential-equations complex-numbers
asked Feb 26 '18 at 16:52
Ewan MillerEwan Miller
12713
$endgroup$
I have a question stating
Find the complementary function of $frac{d^2y}{dx^2} + 4frac{dy}{dx} + 40y = x$
So I solved the auxillary equation, $ lambda^2 + 4lambda +40 = 0$, which gives $lambda = -2 pm 6i$. I believe I am correct in thinking this means the complementary solution is of the form $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, where $x in mathbb{R}$. I would have thought this is the final answer but the answer given is $e^{-2x}(Acos{6x} +Bsin{6x})$. To try to get this I did
$$ begin{align}
Ae^{(-2+6i)x} + Be^{(-2-6i)x} &= Ae^{-2x}(cos{6x} +isin{6x}) +Be^{-2x}(cos{6x}+isin{6x}) \ &= e^{-2x}((A+B)cos{6x} + (A-B)isin{6x})
end{align}$$
How do you get to the given answer?
EDIT
To clarify. What I want to know is how to get from my answer, $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, to the textbook's(and wolfram alpha's) answer, $e^{-2x}(Acos{6x} +Bsin{6x})$ ?
calculus ordinary-differential-equations complex-numbers
calculus ordinary-differential-equations complex-numbers
asked Feb 26 '18 at 16:52
Ewan MillerEwan Miller
12713
asked Feb 26 '18 at 16:52
Ewan MillerEwan Miller
12713
edited Feb 26 '18 at 19:25
Ewan Miller
asked Feb 26 '18 at 16:52
Ewan MillerEwan Miller
12713
asked Feb 26 '18 at 16:52
Ewan MillerEwan Miller
12713
asked Feb 26 '18 at 16:52
Ewan MillerEwan Miller
12713
12713
$begingroup$
You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
$endgroup$
– LutzL
Feb 26 '18 at 17:34
add a comment |
$begingroup$
You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
$endgroup$
– LutzL
Feb 26 '18 at 17:34
$begingroup$
You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
$endgroup$
– LutzL
Feb 26 '18 at 17:34
$begingroup$
You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
$endgroup$
– LutzL
Feb 26 '18 at 17:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is simply $(A+B)=C$ and $(A-B)= D$ to give $e^{-2x}(Ccos{6x}+Dsin{6x})$
answered Feb 26 '18 at 19:59
Ewan MillerEwan Miller
12713
$endgroup$
add a comment |
$begingroup$
It is easy..
Use $e^{ix}=cos x+i sin x$.
And in final step you can collect $cos$ and $sin$ values as constant (you can say $A+B=$ new constant let say $C$ and $(A-B)i=$ another constant let say $D$)
edited Dec 29 '18 at 5:23
Siong Thye Goh
103k1468119
answered Dec 29 '18 at 5:00
محبس الجنمحبس الجن
1
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is simply $(A+B)=C$ and $(A-B)= D$ to give $e^{-2x}(Ccos{6x}+Dsin{6x})$
answered Feb 26 '18 at 19:59
Ewan MillerEwan Miller
12713
$endgroup$
add a comment |
$begingroup$
It is simply $(A+B)=C$ and $(A-B)= D$ to give $e^{-2x}(Ccos{6x}+Dsin{6x})$
answered Feb 26 '18 at 19:59
Ewan MillerEwan Miller
12713
$endgroup$
add a comment |
$begingroup$
It is simply $(A+B)=C$ and $(A-B)= D$ to give $e^{-2x}(Ccos{6x}+Dsin{6x})$
answered Feb 26 '18 at 19:59
Ewan MillerEwan Miller
12713
$endgroup$
It is simply $(A+B)=C$ and $(A-B)= D$ to give $e^{-2x}(Ccos{6x}+Dsin{6x})$
answered Feb 26 '18 at 19:59
Ewan MillerEwan Miller
12713
answered Feb 26 '18 at 19:59
Ewan MillerEwan Miller
12713
answered Feb 26 '18 at 19:59
Ewan MillerEwan Miller
12713
answered Feb 26 '18 at 19:59
Ewan MillerEwan Miller
12713
12713
add a comment |
add a comment |
$begingroup$
It is easy..
Use $e^{ix}=cos x+i sin x$.
And in final step you can collect $cos$ and $sin$ values as constant (you can say $A+B=$ new constant let say $C$ and $(A-B)i=$ another constant let say $D$)
edited Dec 29 '18 at 5:23
Siong Thye Goh
103k1468119
answered Dec 29 '18 at 5:00
محبس الجنمحبس الجن
1
$endgroup$
add a comment |
$begingroup$
It is easy..
Use $e^{ix}=cos x+i sin x$.
And in final step you can collect $cos$ and $sin$ values as constant (you can say $A+B=$ new constant let say $C$ and $(A-B)i=$ another constant let say $D$)
edited Dec 29 '18 at 5:23
Siong Thye Goh
103k1468119
answered Dec 29 '18 at 5:00
محبس الجنمحبس الجن
1
$endgroup$
add a comment |
$begingroup$
It is easy..
Use $e^{ix}=cos x+i sin x$.
And in final step you can collect $cos$ and $sin$ values as constant (you can say $A+B=$ new constant let say $C$ and $(A-B)i=$ another constant let say $D$)
edited Dec 29 '18 at 5:23
Siong Thye Goh
103k1468119
answered Dec 29 '18 at 5:00
محبس الجنمحبس الجن
1
$endgroup$
It is easy..
Use $e^{ix}=cos x+i sin x$.
And in final step you can collect $cos$ and $sin$ values as constant (you can say $A+B=$ new constant let say $C$ and $(A-B)i=$ another constant let say $D$)
edited Dec 29 '18 at 5:23
Siong Thye Goh
103k1468119
answered Dec 29 '18 at 5:00
محبس الجنمحبس الجن
1
edited Dec 29 '18 at 5:23
Siong Thye Goh
103k1468119
edited Dec 29 '18 at 5:23
Siong Thye Goh
103k1468119
edited Dec 29 '18 at 5:23
Siong Thye Goh
103k1468119
103k1468119
answered Dec 29 '18 at 5:00
محبس الجنمحبس الجن
1
answered Dec 29 '18 at 5:00
محبس الجنمحبس الجن
1
answered Dec 29 '18 at 5:00
محبس الجنمحبس الجن
1
1
add a comment |
add a comment |
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$begingroup$
You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
$endgroup$
– LutzL
Feb 26 '18 at 17:34