Given $y = Acos(kt) + Bsin(kt)$, where $A, B, k$ are constants, prove that $y^{(2)} + (k^2)y = 0$












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







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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
















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







share|cite|improve this question











$endgroup$








  • 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














0












0








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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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














  • 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










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 $cos(x)$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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






    share|cite|improve this answer











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      2 Answers
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      2 Answers
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      0












      $begingroup$

      $$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
      You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        $$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
        You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          $$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
          You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.






          share|cite|improve this answer









          $endgroup$



          $$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
          You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 11:16









          Sameer BahetiSameer Baheti

          5168




          5168























              0












              $begingroup$

              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






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                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






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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






                  share|cite|improve this answer











                  $endgroup$



                  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







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 11 at 12:25

























                  answered Dec 3 '18 at 19:38









                  Paras KhoslaParas Khosla

                  699213




                  699213






























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