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
$endgroup$
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
$endgroup$
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
$endgroup$
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
699213
699213
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
active
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 $cos(x)$.
$endgroup$
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
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023555%2fgiven-y-a-coskt-b-sinkt-where-a-b-k-are-constants-prove-that-y%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
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 $cos(x)$.
$endgroup$
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 $cos(x)$.
$endgroup$
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 $cos(x)$.
$endgroup$
$$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
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
5168
5168
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
edited Jan 11 at 12:25
answered Dec 3 '18 at 19:38
Paras KhoslaParas Khosla
699213
699213
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023555%2fgiven-y-a-coskt-b-sinkt-where-a-b-k-are-constants-prove-that-y%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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