Particular solution for $y''+y=sin x$
I'm trying to find the particular solution $y_p$ for:
$$y''+y=sin x$$
I set $y_p(x)=Asin x + Bcos x$ and differentiate 2 times:
$$y_p'(x)=Acos x - Bsin x$$
$$y_p''(x)=-Asin x - Bcos x$$
I Insert into $$y_p''+y_p=sin x$$
$$-Asin x - Bcos x+ Asin x + Bcos x=sin x$$
And this cant be right, because the LHS. is $=0$
Where did I go wrong?
calculus differential-equations
asked Nov 25 '18 at 16:27
Curl
7414
|
show 2 more comments
I'm trying to find the particular solution $y_p$ for:
$$y''+y=sin x$$
I set $y_p(x)=Asin x + Bcos x$ and differentiate 2 times:
$$y_p'(x)=Acos x - Bsin x$$
$$y_p''(x)=-Asin x - Bcos x$$
I Insert into $$y_p''+y_p=sin x$$
$$-Asin x - Bcos x+ Asin x + Bcos x=sin x$$
And this cant be right, because the LHS. is $=0$
Where did I go wrong?
calculus differential-equations
asked Nov 25 '18 at 16:27
Curl
7414
3
Thats because your $y_p$ is solution to the homogenous $y''+y=0$.
– Yadati Kiran
Nov 25 '18 at 16:30
3
The method of undetermined coefficients demands that you add a factor $x$ in the resonance case (with multiplicity one), $y_p(x)=Axsin x+Bxcos x.$
– LutzL
Nov 25 '18 at 16:36
@Curl: In such cases you may take $y_p=Axsin x$ alone (if you are adventurous).
– Yadati Kiran
Nov 25 '18 at 16:48
@LutzL Nice, this works.
– Curl
Nov 25 '18 at 16:48
1
@Isham: That's why I said "if you are adventurous" (i.e. if you realise you can't if you have terms in $cos x$).
– Yadati Kiran
Nov 25 '18 at 17:18
|
show 2 more comments
I'm trying to find the particular solution $y_p$ for:
$$y''+y=sin x$$
I set $y_p(x)=Asin x + Bcos x$ and differentiate 2 times:
$$y_p'(x)=Acos x - Bsin x$$
$$y_p''(x)=-Asin x - Bcos x$$
I Insert into $$y_p''+y_p=sin x$$
$$-Asin x - Bcos x+ Asin x + Bcos x=sin x$$
And this cant be right, because the LHS. is $=0$
Where did I go wrong?
calculus differential-equations
asked Nov 25 '18 at 16:27
Curl
7414
I'm trying to find the particular solution $y_p$ for:
$$y''+y=sin x$$
I set $y_p(x)=Asin x + Bcos x$ and differentiate 2 times:
$$y_p'(x)=Acos x - Bsin x$$
$$y_p''(x)=-Asin x - Bcos x$$
I Insert into $$y_p''+y_p=sin x$$
$$-Asin x - Bcos x+ Asin x + Bcos x=sin x$$
And this cant be right, because the LHS. is $=0$
Where did I go wrong?
calculus differential-equations
calculus differential-equations
asked Nov 25 '18 at 16:27
Curl
7414
asked Nov 25 '18 at 16:27
Curl
7414
asked Nov 25 '18 at 16:27
Curl
7414
asked Nov 25 '18 at 16:27
Curl
7414
asked Nov 25 '18 at 16:27
Curl
7414
7414
3
Thats because your $y_p$ is solution to the homogenous $y''+y=0$.
– Yadati Kiran
Nov 25 '18 at 16:30
3
The method of undetermined coefficients demands that you add a factor $x$ in the resonance case (with multiplicity one), $y_p(x)=Axsin x+Bxcos x.$
– LutzL
Nov 25 '18 at 16:36
@Curl: In such cases you may take $y_p=Axsin x$ alone (if you are adventurous).
– Yadati Kiran
Nov 25 '18 at 16:48
@LutzL Nice, this works.
– Curl
Nov 25 '18 at 16:48
1
@Isham: That's why I said "if you are adventurous" (i.e. if you realise you can't if you have terms in $cos x$).
– Yadati Kiran
Nov 25 '18 at 17:18
|
show 2 more comments
3
Thats because your $y_p$ is solution to the homogenous $y''+y=0$.
– Yadati Kiran
Nov 25 '18 at 16:30
3
The method of undetermined coefficients demands that you add a factor $x$ in the resonance case (with multiplicity one), $y_p(x)=Axsin x+Bxcos x.$
– LutzL
Nov 25 '18 at 16:36
@Curl: In such cases you may take $y_p=Axsin x$ alone (if you are adventurous).
– Yadati Kiran
Nov 25 '18 at 16:48
@LutzL Nice, this works.
– Curl
Nov 25 '18 at 16:48
1
@Isham: That's why I said "if you are adventurous" (i.e. if you realise you can't if you have terms in $cos x$).
– Yadati Kiran
Nov 25 '18 at 17:18
3
3
Thats because your $y_p$ is solution to the homogenous $y''+y=0$.
– Yadati Kiran
Nov 25 '18 at 16:30
Thats because your $y_p$ is solution to the homogenous $y''+y=0$.
– Yadati Kiran
Nov 25 '18 at 16:30
3
3
The method of undetermined coefficients demands that you add a factor $x$ in the resonance case (with multiplicity one), $y_p(x)=Axsin x+Bxcos x.$
– LutzL
Nov 25 '18 at 16:36
The method of undetermined coefficients demands that you add a factor $x$ in the resonance case (with multiplicity one), $y_p(x)=Axsin x+Bxcos x.$
– LutzL
Nov 25 '18 at 16:36
@Curl: In such cases you may take $y_p=Axsin x$ alone (if you are adventurous).
– Yadati Kiran
Nov 25 '18 at 16:48
@Curl: In such cases you may take $y_p=Axsin x$ alone (if you are adventurous).
– Yadati Kiran
Nov 25 '18 at 16:48
@LutzL Nice, this works.
– Curl
Nov 25 '18 at 16:48
@LutzL Nice, this works.
– Curl
Nov 25 '18 at 16:48
1
1
@Isham: That's why I said "if you are adventurous" (i.e. if you realise you can't if you have terms in $cos x$).
– Yadati Kiran
Nov 25 '18 at 17:18
@Isham: That's why I said "if you are adventurous" (i.e. if you realise you can't if you have terms in $cos x$).
– Yadati Kiran
Nov 25 '18 at 17:18
|
show 2 more comments
1 Answer
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When the RHS has the form
$$P_n(x)sin(omega x)$$
and
$(iomega)$ is a root of the caracteristic equation, the particular solution of the equation $$y''+y=P_n(x)sin(omega x)$$
will be as
$$y_p=color{red}{xQ_n(x)}Bigl(Acos(omega x)+Bsin(omega x)Bigr)$$
$P_n$ and $Q_n$ are polynomial of degree $=n$.
answered Nov 25 '18 at 16:53
hamam_Abdallah
37.9k21634
add a comment |
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1 Answer
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When the RHS has the form
$$P_n(x)sin(omega x)$$
and
$(iomega)$ is a root of the caracteristic equation, the particular solution of the equation $$y''+y=P_n(x)sin(omega x)$$
will be as
$$y_p=color{red}{xQ_n(x)}Bigl(Acos(omega x)+Bsin(omega x)Bigr)$$
$P_n$ and $Q_n$ are polynomial of degree $=n$.
answered Nov 25 '18 at 16:53
hamam_Abdallah
37.9k21634
add a comment |
When the RHS has the form
$$P_n(x)sin(omega x)$$
and
$(iomega)$ is a root of the caracteristic equation, the particular solution of the equation $$y''+y=P_n(x)sin(omega x)$$
will be as
$$y_p=color{red}{xQ_n(x)}Bigl(Acos(omega x)+Bsin(omega x)Bigr)$$
$P_n$ and $Q_n$ are polynomial of degree $=n$.
answered Nov 25 '18 at 16:53
hamam_Abdallah
37.9k21634
add a comment |
When the RHS has the form
$$P_n(x)sin(omega x)$$
and
$(iomega)$ is a root of the caracteristic equation, the particular solution of the equation $$y''+y=P_n(x)sin(omega x)$$
will be as
$$y_p=color{red}{xQ_n(x)}Bigl(Acos(omega x)+Bsin(omega x)Bigr)$$
$P_n$ and $Q_n$ are polynomial of degree $=n$.
answered Nov 25 '18 at 16:53
hamam_Abdallah
37.9k21634
When the RHS has the form
$$P_n(x)sin(omega x)$$
and
$(iomega)$ is a root of the caracteristic equation, the particular solution of the equation $$y''+y=P_n(x)sin(omega x)$$
will be as
$$y_p=color{red}{xQ_n(x)}Bigl(Acos(omega x)+Bsin(omega x)Bigr)$$
$P_n$ and $Q_n$ are polynomial of degree $=n$.
answered Nov 25 '18 at 16:53
hamam_Abdallah
37.9k21634
answered Nov 25 '18 at 16:53
hamam_Abdallah
37.9k21634
answered Nov 25 '18 at 16:53
hamam_Abdallah
37.9k21634
answered Nov 25 '18 at 16:53
hamam_Abdallah
37.9k21634
37.9k21634
add a comment |
add a comment |
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3
Thats because your $y_p$ is solution to the homogenous $y''+y=0$.
– Yadati Kiran
Nov 25 '18 at 16:30
3
The method of undetermined coefficients demands that you add a factor $x$ in the resonance case (with multiplicity one), $y_p(x)=Axsin x+Bxcos x.$
– LutzL
Nov 25 '18 at 16:36
@Curl: In such cases you may take $y_p=Axsin x$ alone (if you are adventurous).
– Yadati Kiran
Nov 25 '18 at 16:48
@LutzL Nice, this works.
– Curl
Nov 25 '18 at 16:48
1
@Isham: That's why I said "if you are adventurous" (i.e. if you realise you can't if you have terms in $cos x$).
– Yadati Kiran
Nov 25 '18 at 17:18