Given $y = Acos(kt) + Bsin(kt)$, where $A, B, k$ are constants, prove that $y^{(2)} + (k^2)y = 0$
$begingroup$
EDIT: There was a silly typo, $y^n$ is actually $y''$... sorry everyone!
In that case just finding $y''$ and substituting this and $y$ into the left side will yield the answer 0.
I have noted that $y^n$ refers to the nth derivative.
I found that the 1st, 5th, 9th, ..., $(4m+1)$th derivatives have a sign pattern of - and +, respectively, in front of each trig ratio.
That is, $y^n = -Ak^nsin kt + Bk^ncos kt$.
The 2nd, 6th, 10th, ..., $(4m+2)$th derivates have a sign pattern of - and -.
The 3rd, 7th, 11th, ... $(4m+3)$th derivates have a sign pattern of + and -.
Finally, the 4th, 8th, 12th, ..., $(4m+4)$th derivatives have a sign pattern of + and +.
I am not sure if this is relevant in breaking down the proof into cases...
Thank you in advance for any insight.
calculus derivatives trigonometry proof-writing
edited Dec 3 '18 at 20:33
Paras Khosla
1,704219
asked Dec 3 '18 at 2:58
user424712user424712
12
$endgroup$
add a comment |
$begingroup$
EDIT: There was a silly typo, $y^n$ is actually $y''$... sorry everyone!
In that case just finding $y''$ and substituting this and $y$ into the left side will yield the answer 0.
I have noted that $y^n$ refers to the nth derivative.
I found that the 1st, 5th, 9th, ..., $(4m+1)$th derivatives have a sign pattern of - and +, respectively, in front of each trig ratio.
That is, $y^n = -Ak^nsin kt + Bk^ncos kt$.
The 2nd, 6th, 10th, ..., $(4m+2)$th derivates have a sign pattern of - and -.
The 3rd, 7th, 11th, ... $(4m+3)$th derivates have a sign pattern of + and -.
Finally, the 4th, 8th, 12th, ..., $(4m+4)$th derivatives have a sign pattern of + and +.
I am not sure if this is relevant in breaking down the proof into cases...
Thank you in advance for any insight.
calculus derivatives trigonometry proof-writing
edited Dec 3 '18 at 20:33
Paras Khosla
1,704219
asked Dec 3 '18 at 2:58
user424712user424712
12
$endgroup$
2
$begingroup$
I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
$endgroup$
– WW1
Dec 3 '18 at 3:01
2
$begingroup$
Ah! In that case, I just find the second derivative and sub in!
$endgroup$
– user424712
Dec 3 '18 at 3:02
add a comment |
$begingroup$
EDIT: There was a silly typo, $y^n$ is actually $y''$... sorry everyone!
In that case just finding $y''$ and substituting this and $y$ into the left side will yield the answer 0.
I have noted that $y^n$ refers to the nth derivative.
I found that the 1st, 5th, 9th, ..., $(4m+1)$th derivatives have a sign pattern of - and +, respectively, in front of each trig ratio.
That is, $y^n = -Ak^nsin kt + Bk^ncos kt$.
The 2nd, 6th, 10th, ..., $(4m+2)$th derivates have a sign pattern of - and -.
The 3rd, 7th, 11th, ... $(4m+3)$th derivates have a sign pattern of + and -.
Finally, the 4th, 8th, 12th, ..., $(4m+4)$th derivatives have a sign pattern of + and +.
I am not sure if this is relevant in breaking down the proof into cases...
Thank you in advance for any insight.
calculus derivatives trigonometry proof-writing
edited Dec 3 '18 at 20:33
Paras Khosla
1,704219
asked Dec 3 '18 at 2:58
user424712user424712
12
$endgroup$
EDIT: There was a silly typo, $y^n$ is actually $y''$... sorry everyone!
In that case just finding $y''$ and substituting this and $y$ into the left side will yield the answer 0.
I have noted that $y^n$ refers to the nth derivative.
I found that the 1st, 5th, 9th, ..., $(4m+1)$th derivatives have a sign pattern of - and +, respectively, in front of each trig ratio.
That is, $y^n = -Ak^nsin kt + Bk^ncos kt$.
The 2nd, 6th, 10th, ..., $(4m+2)$th derivates have a sign pattern of - and -.
The 3rd, 7th, 11th, ... $(4m+3)$th derivates have a sign pattern of + and -.
Finally, the 4th, 8th, 12th, ..., $(4m+4)$th derivatives have a sign pattern of + and +.
I am not sure if this is relevant in breaking down the proof into cases...
Thank you in advance for any insight.
calculus derivatives trigonometry proof-writing
calculus derivatives trigonometry proof-writing
edited Dec 3 '18 at 20:33
Paras Khosla
1,704219
asked Dec 3 '18 at 2:58
user424712user424712
12
edited Dec 3 '18 at 20:33
Paras Khosla
1,704219
asked Dec 3 '18 at 2:58
user424712user424712
12
edited Dec 3 '18 at 20:33
Paras Khosla
1,704219
edited Dec 3 '18 at 20:33
Paras Khosla
1,704219
edited Dec 3 '18 at 20:33
Paras Khosla
1,704219
1,704219
asked Dec 3 '18 at 2:58
user424712user424712
12
asked Dec 3 '18 at 2:58
user424712user424712
12
asked Dec 3 '18 at 2:58
user424712user424712
12
12
2
$begingroup$
I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
$endgroup$
– WW1
Dec 3 '18 at 3:01
2
$begingroup$
Ah! In that case, I just find the second derivative and sub in!
$endgroup$
– user424712
Dec 3 '18 at 3:02
add a comment |
2
$begingroup$
I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
$endgroup$
– WW1
Dec 3 '18 at 3:01
2
$begingroup$
Ah! In that case, I just find the second derivative and sub in!
$endgroup$
– user424712
Dec 3 '18 at 3:02
2
2
$begingroup$
I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
$endgroup$
– WW1
Dec 3 '18 at 3:01
$begingroup$
I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
$endgroup$
– WW1
Dec 3 '18 at 3:01
2
2
$begingroup$
Ah! In that case, I just find the second derivative and sub in!
$endgroup$
– user424712
Dec 3 '18 at 3:02
$begingroup$
Ah! In that case, I just find the second derivative and sub in!
$endgroup$
– user424712
Dec 3 '18 at 3:02
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
$$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.
answered Dec 3 '18 at 11:16
Sameer BahetiSameer Baheti
5718
$endgroup$
add a comment |
$begingroup$
We have $y=A cos(kt)+B sin(kt)$ where ${ A, B, k }$ are constants. We are required to prove that $y^{(2)}+(k^2)y=0$ where $y^{(2)}$ denotes the second derivative of $y$ with respect to $t$.
$frac{dy}{dt}=-Ak sin(kt)+Bk cos(kt) implies frac{d^2y}{dt^2}=-Ak^2 cos(kt)+Bk^2 sin(kt) tag1$
Also, $k^2y=Ak^2 cos(kt)+Bk^2 sin(kt) tag2$
Adding equations $(1)$ and $(2)$ together we indeed get $0$ as the result. Hence, Proved
answered Dec 3 '18 at 19:38
Paras KhoslaParas Khosla
1,704219
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.
answered Dec 3 '18 at 11:16
Sameer BahetiSameer Baheti
5718
$endgroup$
add a comment |
$begingroup$
$$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.
answered Dec 3 '18 at 11:16
Sameer BahetiSameer Baheti
5718
$endgroup$
add a comment |
$begingroup$
$$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.
answered Dec 3 '18 at 11:16
Sameer BahetiSameer Baheti
5718
$endgroup$
$$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.
answered Dec 3 '18 at 11:16
Sameer BahetiSameer Baheti
5718
answered Dec 3 '18 at 11:16
Sameer BahetiSameer Baheti
5718
answered Dec 3 '18 at 11:16
Sameer BahetiSameer Baheti
5718
answered Dec 3 '18 at 11:16
Sameer BahetiSameer Baheti
5718
5718
add a comment |
add a comment |
$begingroup$
We have $y=A cos(kt)+B sin(kt)$ where ${ A, B, k }$ are constants. We are required to prove that $y^{(2)}+(k^2)y=0$ where $y^{(2)}$ denotes the second derivative of $y$ with respect to $t$.
$frac{dy}{dt}=-Ak sin(kt)+Bk cos(kt) implies frac{d^2y}{dt^2}=-Ak^2 cos(kt)+Bk^2 sin(kt) tag1$
Also, $k^2y=Ak^2 cos(kt)+Bk^2 sin(kt) tag2$
Adding equations $(1)$ and $(2)$ together we indeed get $0$ as the result. Hence, Proved
answered Dec 3 '18 at 19:38
Paras KhoslaParas Khosla
1,704219
$endgroup$
add a comment |
$begingroup$
We have $y=A cos(kt)+B sin(kt)$ where ${ A, B, k }$ are constants. We are required to prove that $y^{(2)}+(k^2)y=0$ where $y^{(2)}$ denotes the second derivative of $y$ with respect to $t$.
$frac{dy}{dt}=-Ak sin(kt)+Bk cos(kt) implies frac{d^2y}{dt^2}=-Ak^2 cos(kt)+Bk^2 sin(kt) tag1$
Also, $k^2y=Ak^2 cos(kt)+Bk^2 sin(kt) tag2$
Adding equations $(1)$ and $(2)$ together we indeed get $0$ as the result. Hence, Proved
answered Dec 3 '18 at 19:38
Paras KhoslaParas Khosla
1,704219
$endgroup$
add a comment |
$begingroup$
We have $y=A cos(kt)+B sin(kt)$ where ${ A, B, k }$ are constants. We are required to prove that $y^{(2)}+(k^2)y=0$ where $y^{(2)}$ denotes the second derivative of $y$ with respect to $t$.
$frac{dy}{dt}=-Ak sin(kt)+Bk cos(kt) implies frac{d^2y}{dt^2}=-Ak^2 cos(kt)+Bk^2 sin(kt) tag1$
Also, $k^2y=Ak^2 cos(kt)+Bk^2 sin(kt) tag2$
Adding equations $(1)$ and $(2)$ together we indeed get $0$ as the result. Hence, Proved
answered Dec 3 '18 at 19:38
Paras KhoslaParas Khosla
1,704219
$endgroup$
We have $y=A cos(kt)+B sin(kt)$ where ${ A, B, k }$ are constants. We are required to prove that $y^{(2)}+(k^2)y=0$ where $y^{(2)}$ denotes the second derivative of $y$ with respect to $t$.
$frac{dy}{dt}=-Ak sin(kt)+Bk cos(kt) implies frac{d^2y}{dt^2}=-Ak^2 cos(kt)+Bk^2 sin(kt) tag1$
Also, $k^2y=Ak^2 cos(kt)+Bk^2 sin(kt) tag2$
Adding equations $(1)$ and $(2)$ together we indeed get $0$ as the result. Hence, Proved
answered Dec 3 '18 at 19:38
Paras KhoslaParas Khosla
1,704219
edited Jan 11 at 12:25
answered Dec 3 '18 at 19:38
Paras KhoslaParas Khosla
1,704219
answered Dec 3 '18 at 19:38
Paras KhoslaParas Khosla
1,704219
answered Dec 3 '18 at 19:38
Paras KhoslaParas Khosla
1,704219
1,704219
add a comment |
add a comment |
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$begingroup$
I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
$endgroup$
– WW1
Dec 3 '18 at 3:01
2
$begingroup$
Ah! In that case, I just find the second derivative and sub in!
$endgroup$
– user424712
Dec 3 '18 at 3:02