Resolution of differential equation [closed]
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All functions verify the condition below :
$$f''(x)+f(-x)=x+cos(x)$$
I have to use something related to differential equations.
real-analysis differential-equations
edited Dec 13 at 22:57
user376343
2,7782822
asked Dec 13 at 22:50
ahmed bennani
175
closed as off-topic by Saad, KReiser, T. Bongers, Andrés E. Caicedo, BigbearZzz Dec 14 at 5:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, T. Bongers
add a comment |
All functions verify the condition below :
$$f''(x)+f(-x)=x+cos(x)$$
I have to use something related to differential equations.
real-analysis differential-equations
edited Dec 13 at 22:57
user376343
2,7782822
asked Dec 13 at 22:50
ahmed bennani
175
closed as off-topic by Saad, KReiser, T. Bongers, Andrés E. Caicedo, BigbearZzz Dec 14 at 5:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, T. Bongers
add a comment |
All functions verify the condition below :
$$f''(x)+f(-x)=x+cos(x)$$
I have to use something related to differential equations.
real-analysis differential-equations
edited Dec 13 at 22:57
user376343
2,7782822
asked Dec 13 at 22:50
ahmed bennani
175
All functions verify the condition below :
$$f''(x)+f(-x)=x+cos(x)$$
I have to use something related to differential equations.
real-analysis differential-equations
real-analysis differential-equations
edited Dec 13 at 22:57
user376343
2,7782822
asked Dec 13 at 22:50
ahmed bennani
175
edited Dec 13 at 22:57
user376343
2,7782822
asked Dec 13 at 22:50
ahmed bennani
175
edited Dec 13 at 22:57
user376343
2,7782822
edited Dec 13 at 22:57
user376343
2,7782822
edited Dec 13 at 22:57
user376343
2,7782822
2,7782822
asked Dec 13 at 22:50
ahmed bennani
175
asked Dec 13 at 22:50
ahmed bennani
175
asked Dec 13 at 22:50
ahmed bennani
175
175
closed as off-topic by Saad, KReiser, T. Bongers, Andrés E. Caicedo, BigbearZzz Dec 14 at 5:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, T. Bongers
closed as off-topic by Saad, KReiser, T. Bongers, Andrés E. Caicedo, BigbearZzz Dec 14 at 5:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, T. Bongers
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Any real function can be written in a unique way as the sum of an even function and an odd function, say
$$f=g+h$$
where $g$ is even and $h$ is odd. It is easy to check that $g''$ is also even and $h''$ is also odd. So we have
$$g''(x)+h''(x)+g(x)-h(x)=x+cos x .$$
Equating even and odd parts,
$$g''(x)+g(x)=cos x ,qquad h''(x)-h(x)=x .$$
Solving by standard methods,
$$g(x)=Acos x+Bsin x+tfrac12xsin x ,qquad h(x)=Ccosh x+Dsinh x-x .$$
But $g$ is even so $B=0$, and $h$ is odd so $C=0$. Hence
$$f(x)=g(x)+h(x)=Acos x+Dsinh x+tfrac12xsin x-x .$$
answered Dec 13 at 23:13
David
67.7k663126
add a comment |
Iterating the equation you get $f^(4)(x)-f(x)=x-2cos , x$ which can be solved by standard methods. Solutions of $lambda^4-1=0$ are $pm 1, pm i$, so the general solution to $f^(4)(x)-f(x)=0$ is $ae^x+be^-x+ccos , x +dsin ,x$. You will have to solve the inhomogeneous equation and go back to the original DE to see which of these are actually solutions. I hope you can fill in the details.
answered Dec 13 at 23:17
Kavi Rama Murthy
48.9k31854
1
This introduces extra solutions which are not solutions of the given equation.
– David
Dec 13 at 23:26
You get $f^IV(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^IV(x)-f(x)=x-2cos(x),$$ which is different from what you state.
– LutzL
Dec 14 at 1:28
@LutzL You are right. I have revised my answer.
– Kavi Rama Murthy
Dec 14 at 5:30
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Any real function can be written in a unique way as the sum of an even function and an odd function, say
$$f=g+h$$
where $g$ is even and $h$ is odd. It is easy to check that $g''$ is also even and $h''$ is also odd. So we have
$$g''(x)+h''(x)+g(x)-h(x)=x+cos x .$$
Equating even and odd parts,
$$g''(x)+g(x)=cos x ,qquad h''(x)-h(x)=x .$$
Solving by standard methods,
$$g(x)=Acos x+Bsin x+tfrac12xsin x ,qquad h(x)=Ccosh x+Dsinh x-x .$$
But $g$ is even so $B=0$, and $h$ is odd so $C=0$. Hence
$$f(x)=g(x)+h(x)=Acos x+Dsinh x+tfrac12xsin x-x .$$
answered Dec 13 at 23:13
David
67.7k663126
add a comment |
Any real function can be written in a unique way as the sum of an even function and an odd function, say
$$f=g+h$$
where $g$ is even and $h$ is odd. It is easy to check that $g''$ is also even and $h''$ is also odd. So we have
$$g''(x)+h''(x)+g(x)-h(x)=x+cos x .$$
Equating even and odd parts,
$$g''(x)+g(x)=cos x ,qquad h''(x)-h(x)=x .$$
Solving by standard methods,
$$g(x)=Acos x+Bsin x+tfrac12xsin x ,qquad h(x)=Ccosh x+Dsinh x-x .$$
But $g$ is even so $B=0$, and $h$ is odd so $C=0$. Hence
$$f(x)=g(x)+h(x)=Acos x+Dsinh x+tfrac12xsin x-x .$$
answered Dec 13 at 23:13
David
67.7k663126
add a comment |
Any real function can be written in a unique way as the sum of an even function and an odd function, say
$$f=g+h$$
where $g$ is even and $h$ is odd. It is easy to check that $g''$ is also even and $h''$ is also odd. So we have
$$g''(x)+h''(x)+g(x)-h(x)=x+cos x .$$
Equating even and odd parts,
$$g''(x)+g(x)=cos x ,qquad h''(x)-h(x)=x .$$
Solving by standard methods,
$$g(x)=Acos x+Bsin x+tfrac12xsin x ,qquad h(x)=Ccosh x+Dsinh x-x .$$
But $g$ is even so $B=0$, and $h$ is odd so $C=0$. Hence
$$f(x)=g(x)+h(x)=Acos x+Dsinh x+tfrac12xsin x-x .$$
answered Dec 13 at 23:13
David
67.7k663126
Any real function can be written in a unique way as the sum of an even function and an odd function, say
$$f=g+h$$
where $g$ is even and $h$ is odd. It is easy to check that $g''$ is also even and $h''$ is also odd. So we have
$$g''(x)+h''(x)+g(x)-h(x)=x+cos x .$$
Equating even and odd parts,
$$g''(x)+g(x)=cos x ,qquad h''(x)-h(x)=x .$$
Solving by standard methods,
$$g(x)=Acos x+Bsin x+tfrac12xsin x ,qquad h(x)=Ccosh x+Dsinh x-x .$$
But $g$ is even so $B=0$, and $h$ is odd so $C=0$. Hence
$$f(x)=g(x)+h(x)=Acos x+Dsinh x+tfrac12xsin x-x .$$
answered Dec 13 at 23:13
David
67.7k663126
answered Dec 13 at 23:13
David
67.7k663126
answered Dec 13 at 23:13
David
67.7k663126
answered Dec 13 at 23:13
David
67.7k663126
67.7k663126
add a comment |
add a comment |
Iterating the equation you get $f^(4)(x)-f(x)=x-2cos , x$ which can be solved by standard methods. Solutions of $lambda^4-1=0$ are $pm 1, pm i$, so the general solution to $f^(4)(x)-f(x)=0$ is $ae^x+be^-x+ccos , x +dsin ,x$. You will have to solve the inhomogeneous equation and go back to the original DE to see which of these are actually solutions. I hope you can fill in the details.
answered Dec 13 at 23:17
Kavi Rama Murthy
48.9k31854
1
This introduces extra solutions which are not solutions of the given equation.
– David
Dec 13 at 23:26
You get $f^IV(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^IV(x)-f(x)=x-2cos(x),$$ which is different from what you state.
– LutzL
Dec 14 at 1:28
@LutzL You are right. I have revised my answer.
– Kavi Rama Murthy
Dec 14 at 5:30
add a comment |
Iterating the equation you get $f^(4)(x)-f(x)=x-2cos , x$ which can be solved by standard methods. Solutions of $lambda^4-1=0$ are $pm 1, pm i$, so the general solution to $f^(4)(x)-f(x)=0$ is $ae^x+be^-x+ccos , x +dsin ,x$. You will have to solve the inhomogeneous equation and go back to the original DE to see which of these are actually solutions. I hope you can fill in the details.
answered Dec 13 at 23:17
Kavi Rama Murthy
48.9k31854
1
This introduces extra solutions which are not solutions of the given equation.
– David
Dec 13 at 23:26
You get $f^IV(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^IV(x)-f(x)=x-2cos(x),$$ which is different from what you state.
– LutzL
Dec 14 at 1:28
@LutzL You are right. I have revised my answer.
– Kavi Rama Murthy
Dec 14 at 5:30
add a comment |
Iterating the equation you get $f^(4)(x)-f(x)=x-2cos , x$ which can be solved by standard methods. Solutions of $lambda^4-1=0$ are $pm 1, pm i$, so the general solution to $f^(4)(x)-f(x)=0$ is $ae^x+be^-x+ccos , x +dsin ,x$. You will have to solve the inhomogeneous equation and go back to the original DE to see which of these are actually solutions. I hope you can fill in the details.
answered Dec 13 at 23:17
Kavi Rama Murthy
48.9k31854
Iterating the equation you get $f^(4)(x)-f(x)=x-2cos , x$ which can be solved by standard methods. Solutions of $lambda^4-1=0$ are $pm 1, pm i$, so the general solution to $f^(4)(x)-f(x)=0$ is $ae^x+be^-x+ccos , x +dsin ,x$. You will have to solve the inhomogeneous equation and go back to the original DE to see which of these are actually solutions. I hope you can fill in the details.
answered Dec 13 at 23:17
Kavi Rama Murthy
48.9k31854
edited Dec 14 at 5:29
answered Dec 13 at 23:17
Kavi Rama Murthy
48.9k31854
answered Dec 13 at 23:17
Kavi Rama Murthy
48.9k31854
answered Dec 13 at 23:17
Kavi Rama Murthy
48.9k31854
48.9k31854
1
This introduces extra solutions which are not solutions of the given equation.
– David
Dec 13 at 23:26
You get $f^IV(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^IV(x)-f(x)=x-2cos(x),$$ which is different from what you state.
– LutzL
Dec 14 at 1:28
@LutzL You are right. I have revised my answer.
– Kavi Rama Murthy
Dec 14 at 5:30
add a comment |
1
This introduces extra solutions which are not solutions of the given equation.
– David
Dec 13 at 23:26
You get $f^IV(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^IV(x)-f(x)=x-2cos(x),$$ which is different from what you state.
– LutzL
Dec 14 at 1:28
@LutzL You are right. I have revised my answer.
– Kavi Rama Murthy
Dec 14 at 5:30
1
1
This introduces extra solutions which are not solutions of the given equation.
– David
Dec 13 at 23:26
This introduces extra solutions which are not solutions of the given equation.
– David
Dec 13 at 23:26
You get $f^IV(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^IV(x)-f(x)=x-2cos(x),$$ which is different from what you state.
– LutzL
Dec 14 at 1:28
You get $f^IV(x)+f''(-x)=-cos(x)$ and $f''(-x)+f(x)=-x+cos(x)$, therefore $$f^IV(x)-f(x)=x-2cos(x),$$ which is different from what you state.
– LutzL
Dec 14 at 1:28
@LutzL You are right. I have revised my answer.
– Kavi Rama Murthy
Dec 14 at 5:30
@LutzL You are right. I have revised my answer.
– Kavi Rama Murthy
Dec 14 at 5:30
add a comment |
