Particular solutions to $y''+4y=cos^2(x)$
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?
differential-equations
edited Jan 4 at 14:52
zipirovich
11.2k11631
asked Jan 4 at 14:38
AlexisAlexis
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add a comment |
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?
differential-equations
edited Jan 4 at 14:52
zipirovich
11.2k11631
asked Jan 4 at 14:38
AlexisAlexis
82
New contributor
Alexis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
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?
– Oscar Lanzi
Jan 4 at 14:46
No I can't find out the way to get a particular solution
– Alexis
Jan 4 at 14:55
add a comment |
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?
differential-equations
edited Jan 4 at 14:52
zipirovich
11.2k11631
asked Jan 4 at 14:38
AlexisAlexis
82
New contributor
Alexis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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?
differential-equations
differential-equations
edited Jan 4 at 14:52
zipirovich
11.2k11631
asked Jan 4 at 14:38
AlexisAlexis
82
New contributor
Alexis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 14:52
zipirovich
11.2k11631
asked Jan 4 at 14:38
AlexisAlexis
82
New contributor
Alexis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 14:52
zipirovich
11.2k11631
edited Jan 4 at 14:52
zipirovich
11.2k11631
edited Jan 4 at 14:52
zipirovich
11.2k11631
11.2k11631
asked Jan 4 at 14:38
AlexisAlexis
82
New contributor
Alexis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 14:38
AlexisAlexis
82
asked Jan 4 at 14:38
AlexisAlexis
82
82
New contributor
Alexis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alexis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alexis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
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?
– Oscar Lanzi
Jan 4 at 14:46
No I can't find out the way to get a particular solution
– Alexis
Jan 4 at 14:55
add a comment |
1
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?
– Oscar Lanzi
Jan 4 at 14:46
No I can't find out the way to get a particular solution
– Alexis
Jan 4 at 14:55
1
1
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?
– Oscar Lanzi
Jan 4 at 14:46
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?
– Oscar Lanzi
Jan 4 at 14:46
No I can't find out the way to get a particular solution
– Alexis
Jan 4 at 14:55
No I can't find out the way to get a particular solution
– Alexis
Jan 4 at 14:55
add a comment |
1 Answer
1
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votes
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 $y_{p2}(x)=A$.
answered Jan 4 at 15:12
zipirovichzipirovich
11.2k11631
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 ?
– Alexis
Jan 4 at 15:19
@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 $y_p=y_{p1}+y_{p2}$.
– zipirovich
Jan 4 at 15:21
Alright then I get it thank you very much for your help !
– Alexis
Jan 4 at 15:25
add a comment |
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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 $y_{p2}(x)=A$.
answered Jan 4 at 15:12
zipirovichzipirovich
11.2k11631
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 ?
– Alexis
Jan 4 at 15:19
@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 $y_p=y_{p1}+y_{p2}$.
– zipirovich
Jan 4 at 15:21
Alright then I get it thank you very much for your help !
– Alexis
Jan 4 at 15:25
add a comment |
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 $y_{p2}(x)=A$.
answered Jan 4 at 15:12
zipirovichzipirovich
11.2k11631
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 ?
– Alexis
Jan 4 at 15:19
@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 $y_p=y_{p1}+y_{p2}$.
– zipirovich
Jan 4 at 15:21
Alright then I get it thank you very much for your help !
– Alexis
Jan 4 at 15:25
add a comment |
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 $y_{p2}(x)=A$.
answered Jan 4 at 15:12
zipirovichzipirovich
11.2k11631
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 $y_{p2}(x)=A$.
answered Jan 4 at 15:12
zipirovichzipirovich
11.2k11631
answered Jan 4 at 15:12
zipirovichzipirovich
11.2k11631
answered Jan 4 at 15:12
zipirovichzipirovich
11.2k11631
answered Jan 4 at 15:12
zipirovichzipirovich
11.2k11631
11.2k11631
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 ?
– Alexis
Jan 4 at 15:19
@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 $y_p=y_{p1}+y_{p2}$.
– zipirovich
Jan 4 at 15:21
Alright then I get it thank you very much for your help !
– Alexis
Jan 4 at 15:25
add a comment |
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 ?
– Alexis
Jan 4 at 15:19
@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 $y_p=y_{p1}+y_{p2}$.
– zipirovich
Jan 4 at 15:21
Alright then I get it thank you very much for your help !
– Alexis
Jan 4 at 15:25
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 ?
– Alexis
Jan 4 at 15:19
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 ?
– Alexis
Jan 4 at 15:19
@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 $y_p=y_{p1}+y_{p2}$.
– zipirovich
Jan 4 at 15:21
@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 $y_p=y_{p1}+y_{p2}$.
– zipirovich
Jan 4 at 15:21
Alright then I get it thank you very much for your help !
– Alexis
Jan 4 at 15:25
Alright then I get it thank you very much for your help !
– Alexis
Jan 4 at 15:25
add a comment |
Alexis is a new contributor. Be nice, and check out our Code of Conduct.
Alexis is a new contributor. Be nice, and check out our Code of Conduct.
Alexis is a new contributor. Be nice, and check out our Code of Conduct.
Alexis is a new contributor. Be nice, and check out our Code of Conduct.
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1
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?
– Oscar Lanzi
Jan 4 at 14:46
No I can't find out the way to get a particular solution
– Alexis
Jan 4 at 14:55