Particular solutions to $y''+4y=cos^2(x)$












1












$begingroup$


I have to solve the equation above but after I have solved the homogeneous part:
$$y_h=Acos(2x)+Bsin(2x), ; A,Binmathbb{C},$$
I don't get it how to find $y_p$. I am told to deduct it from $cos^2(x)=frac{1}{2}cos(2x)+frac{1}{2}$, but I don't understand how. Can you help me tho?






ordinary-differential-equations







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$endgroup$








  • 1




    $begingroup$
    Note that $cos 2x$ solves the homogeneous equation and also appears in the nonhomogeneity term. Do you know how to modify it to get a particular solution in such a case?
    $endgroup$
    – Oscar Lanzi
    Jan 4 at 14:46












  • $begingroup$
    No I can't find out the way to get a particular solution
    $endgroup$
    – Alexis
    Jan 4 at 14:55
















1












$begingroup$


I have to solve the equation above but after I have solved the homogeneous part:
$$y_h=Acos(2x)+Bsin(2x), ; A,Binmathbb{C},$$
I don't get it how to find $y_p$. I am told to deduct it from $cos^2(x)=frac{1}{2}cos(2x)+frac{1}{2}$, but I don't understand how. Can you help me tho?






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $cos 2x$ solves the homogeneous equation and also appears in the nonhomogeneity term. Do you know how to modify it to get a particular solution in such a case?
    $endgroup$
    – Oscar Lanzi
    Jan 4 at 14:46












  • $begingroup$
    No I can't find out the way to get a particular solution
    $endgroup$
    – Alexis
    Jan 4 at 14:55














1












1








1


1



$begingroup$


I have to solve the equation above but after I have solved the homogeneous part:
$$y_h=Acos(2x)+Bsin(2x), ; A,Binmathbb{C},$$
I don't get it how to find $y_p$. I am told to deduct it from $cos^2(x)=frac{1}{2}cos(2x)+frac{1}{2}$, but I don't understand how. Can you help me tho?






ordinary-differential-equations







share|cite|improve this question











$endgroup$




I have to solve the equation above but after I have solved the homogeneous part:
$$y_h=Acos(2x)+Bsin(2x), ; A,Binmathbb{C},$$
I don't get it how to find $y_p$. I am told to deduct it from $cos^2(x)=frac{1}{2}cos(2x)+frac{1}{2}$, but I don't understand how. Can you help me tho?






ordinary-differential-equations




ordinary-differential-equations






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 14:52









zipirovich

11.3k11731




11.3k11731










asked Jan 4 at 14:38









AlexisAlexis

82




82








  • 1




    $begingroup$
    Note that $cos 2x$ solves the homogeneous equation and also appears in the nonhomogeneity term. Do you know how to modify it to get a particular solution in such a case?
    $endgroup$
    – Oscar Lanzi
    Jan 4 at 14:46












  • $begingroup$
    No I can't find out the way to get a particular solution
    $endgroup$
    – Alexis
    Jan 4 at 14:55














  • 1




    $begingroup$
    Note that $cos 2x$ solves the homogeneous equation and also appears in the nonhomogeneity term. Do you know how to modify it to get a particular solution in such a case?
    $endgroup$
    – Oscar Lanzi
    Jan 4 at 14:46












  • $begingroup$
    No I can't find out the way to get a particular solution
    $endgroup$
    – Alexis
    Jan 4 at 14:55








1




1




$begingroup$
Note that $cos 2x$ solves the homogeneous equation and also appears in the nonhomogeneity term. Do you know how to modify it to get a particular solution in such a case?
$endgroup$
– Oscar Lanzi
Jan 4 at 14:46






$begingroup$
Note that $cos 2x$ solves the homogeneous equation and also appears in the nonhomogeneity term. Do you know how to modify it to get a particular solution in such a case?
$endgroup$
– Oscar Lanzi
Jan 4 at 14:46














$begingroup$
No I can't find out the way to get a particular solution
$endgroup$
– Alexis
Jan 4 at 14:55




$begingroup$
No I can't find out the way to get a particular solution
$endgroup$
– Alexis
Jan 4 at 14:55










1 Answer
1






active

oldest

votes


















2












$begingroup$

What the hint means is that first of all you literally replace the right-hand side of the equation with $cos^2(x)=frac{1}{2}cos(2x)+frac{1}{2}$, so now you're looking for a particular solution to
$$y''+4y=frac{1}{2}cos(2x)+frac{1}{2}.$$
Moreover, by the superposition principle for particular solutions (see e.g. here or here), we may seek separate solutions to
$$y''+4y=frac{1}{2}cos(2x) quad text{and} quad y''+4y=frac{1}{2}$$
and then add them together.




  • For the first equation $y''+4y=frac{1}{2}cos(2x)$, note that $lambda=pm2i$ were the roots of the characteristic equation, so you should set up a particular solution in the form $y_{p1}(x)=x(Acos(2x)+Bsin(2x))$.

  • For the second equation $y''+4y=frac{1}{2}$, note that $lambda=0$ was not a root of the characteristic equation, so you should set up a particular solution in the form $y_{p2}(x)=A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok I think I see but with the principle of superposition shouldn't I find 2 particular solutions of each new ODE as it is a 2nd order ?
    $endgroup$
    – Alexis
    Jan 4 at 15:19












  • $begingroup$
    @Alexis: Yes, that's exactly what I said. You find two particular solutions, so to speak. More precisely, you find one particular solution to each of the separate non-homogeneous equations: $y_{p1}$ for the first and $y_{p2}$ for the second. Then a particular solution to the original non-homogeneous equation will be $y_p=y_{p1}+y_{p2}$.
    $endgroup$
    – zipirovich
    Jan 4 at 15:21










  • $begingroup$
    Alright then I get it thank you very much for your help !
    $endgroup$
    – Alexis
    Jan 4 at 15:25














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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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2












$begingroup$

What the hint means is that first of all you literally replace the right-hand side of the equation with $cos^2(x)=frac{1}{2}cos(2x)+frac{1}{2}$, so now you're looking for a particular solution to
$$y''+4y=frac{1}{2}cos(2x)+frac{1}{2}.$$
Moreover, by the superposition principle for particular solutions (see e.g. here or here), we may seek separate solutions to
$$y''+4y=frac{1}{2}cos(2x) quad text{and} quad y''+4y=frac{1}{2}$$
and then add them together.




  • For the first equation $y''+4y=frac{1}{2}cos(2x)$, note that $lambda=pm2i$ were the roots of the characteristic equation, so you should set up a particular solution in the form $y_{p1}(x)=x(Acos(2x)+Bsin(2x))$.

  • For the second equation $y''+4y=frac{1}{2}$, note that $lambda=0$ was not a root of the characteristic equation, so you should set up a particular solution in the form $y_{p2}(x)=A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok I think I see but with the principle of superposition shouldn't I find 2 particular solutions of each new ODE as it is a 2nd order ?
    $endgroup$
    – Alexis
    Jan 4 at 15:19












  • $begingroup$
    @Alexis: Yes, that's exactly what I said. You find two particular solutions, so to speak. More precisely, you find one particular solution to each of the separate non-homogeneous equations: $y_{p1}$ for the first and $y_{p2}$ for the second. Then a particular solution to the original non-homogeneous equation will be $y_p=y_{p1}+y_{p2}$.
    $endgroup$
    – zipirovich
    Jan 4 at 15:21










  • $begingroup$
    Alright then I get it thank you very much for your help !
    $endgroup$
    – Alexis
    Jan 4 at 15:25


















2












$begingroup$

What the hint means is that first of all you literally replace the right-hand side of the equation with $cos^2(x)=frac{1}{2}cos(2x)+frac{1}{2}$, so now you're looking for a particular solution to
$$y''+4y=frac{1}{2}cos(2x)+frac{1}{2}.$$
Moreover, by the superposition principle for particular solutions (see e.g. here or here), we may seek separate solutions to
$$y''+4y=frac{1}{2}cos(2x) quad text{and} quad y''+4y=frac{1}{2}$$
and then add them together.




  • For the first equation $y''+4y=frac{1}{2}cos(2x)$, note that $lambda=pm2i$ were the roots of the characteristic equation, so you should set up a particular solution in the form $y_{p1}(x)=x(Acos(2x)+Bsin(2x))$.

  • For the second equation $y''+4y=frac{1}{2}$, note that $lambda=0$ was not a root of the characteristic equation, so you should set up a particular solution in the form $y_{p2}(x)=A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok I think I see but with the principle of superposition shouldn't I find 2 particular solutions of each new ODE as it is a 2nd order ?
    $endgroup$
    – Alexis
    Jan 4 at 15:19












  • $begingroup$
    @Alexis: Yes, that's exactly what I said. You find two particular solutions, so to speak. More precisely, you find one particular solution to each of the separate non-homogeneous equations: $y_{p1}$ for the first and $y_{p2}$ for the second. Then a particular solution to the original non-homogeneous equation will be $y_p=y_{p1}+y_{p2}$.
    $endgroup$
    – zipirovich
    Jan 4 at 15:21










  • $begingroup$
    Alright then I get it thank you very much for your help !
    $endgroup$
    – Alexis
    Jan 4 at 15:25
















2












2








2





$begingroup$

What the hint means is that first of all you literally replace the right-hand side of the equation with $cos^2(x)=frac{1}{2}cos(2x)+frac{1}{2}$, so now you're looking for a particular solution to
$$y''+4y=frac{1}{2}cos(2x)+frac{1}{2}.$$
Moreover, by the superposition principle for particular solutions (see e.g. here or here), we may seek separate solutions to
$$y''+4y=frac{1}{2}cos(2x) quad text{and} quad y''+4y=frac{1}{2}$$
and then add them together.




  • For the first equation $y''+4y=frac{1}{2}cos(2x)$, note that $lambda=pm2i$ were the roots of the characteristic equation, so you should set up a particular solution in the form $y_{p1}(x)=x(Acos(2x)+Bsin(2x))$.

  • For the second equation $y''+4y=frac{1}{2}$, note that $lambda=0$ was not a root of the characteristic equation, so you should set up a particular solution in the form $y_{p2}(x)=A$.






share|cite|improve this answer









$endgroup$



What the hint means is that first of all you literally replace the right-hand side of the equation with $cos^2(x)=frac{1}{2}cos(2x)+frac{1}{2}$, so now you're looking for a particular solution to
$$y''+4y=frac{1}{2}cos(2x)+frac{1}{2}.$$
Moreover, by the superposition principle for particular solutions (see e.g. here or here), we may seek separate solutions to
$$y''+4y=frac{1}{2}cos(2x) quad text{and} quad y''+4y=frac{1}{2}$$
and then add them together.




  • For the first equation $y''+4y=frac{1}{2}cos(2x)$, note that $lambda=pm2i$ were the roots of the characteristic equation, so you should set up a particular solution in the form $y_{p1}(x)=x(Acos(2x)+Bsin(2x))$.

  • For the second equation $y''+4y=frac{1}{2}$, note that $lambda=0$ was not a root of the characteristic equation, so you should set up a particular solution in the form $y_{p2}(x)=A$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 15:12









zipirovichzipirovich

11.3k11731




11.3k11731












  • $begingroup$
    Ok I think I see but with the principle of superposition shouldn't I find 2 particular solutions of each new ODE as it is a 2nd order ?
    $endgroup$
    – Alexis
    Jan 4 at 15:19












  • $begingroup$
    @Alexis: Yes, that's exactly what I said. You find two particular solutions, so to speak. More precisely, you find one particular solution to each of the separate non-homogeneous equations: $y_{p1}$ for the first and $y_{p2}$ for the second. Then a particular solution to the original non-homogeneous equation will be $y_p=y_{p1}+y_{p2}$.
    $endgroup$
    – zipirovich
    Jan 4 at 15:21










  • $begingroup$
    Alright then I get it thank you very much for your help !
    $endgroup$
    – Alexis
    Jan 4 at 15:25




















  • $begingroup$
    Ok I think I see but with the principle of superposition shouldn't I find 2 particular solutions of each new ODE as it is a 2nd order ?
    $endgroup$
    – Alexis
    Jan 4 at 15:19












  • $begingroup$
    @Alexis: Yes, that's exactly what I said. You find two particular solutions, so to speak. More precisely, you find one particular solution to each of the separate non-homogeneous equations: $y_{p1}$ for the first and $y_{p2}$ for the second. Then a particular solution to the original non-homogeneous equation will be $y_p=y_{p1}+y_{p2}$.
    $endgroup$
    – zipirovich
    Jan 4 at 15:21










  • $begingroup$
    Alright then I get it thank you very much for your help !
    $endgroup$
    – Alexis
    Jan 4 at 15:25


















$begingroup$
Ok I think I see but with the principle of superposition shouldn't I find 2 particular solutions of each new ODE as it is a 2nd order ?
$endgroup$
– Alexis
Jan 4 at 15:19






$begingroup$
Ok I think I see but with the principle of superposition shouldn't I find 2 particular solutions of each new ODE as it is a 2nd order ?
$endgroup$
– Alexis
Jan 4 at 15:19














$begingroup$
@Alexis: Yes, that's exactly what I said. You find two particular solutions, so to speak. More precisely, you find one particular solution to each of the separate non-homogeneous equations: $y_{p1}$ for the first and $y_{p2}$ for the second. Then a particular solution to the original non-homogeneous equation will be $y_p=y_{p1}+y_{p2}$.
$endgroup$
– zipirovich
Jan 4 at 15:21




$begingroup$
@Alexis: Yes, that's exactly what I said. You find two particular solutions, so to speak. More precisely, you find one particular solution to each of the separate non-homogeneous equations: $y_{p1}$ for the first and $y_{p2}$ for the second. Then a particular solution to the original non-homogeneous equation will be $y_p=y_{p1}+y_{p2}$.
$endgroup$
– zipirovich
Jan 4 at 15:21












$begingroup$
Alright then I get it thank you very much for your help !
$endgroup$
– Alexis
Jan 4 at 15:25






$begingroup$
Alright then I get it thank you very much for your help !
$endgroup$
– Alexis
Jan 4 at 15:25




















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