particular solution to $y^4 -2y'' +y = xe^x $

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1
down vote

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I am trying to find the particular solution to the $y^4 -2y'' +y = xe^x $
and currently am misunderstanding what to do.

My steps:

the polynomial operator concerned is $p(s)= s^4 -2s^2 + 1 $
which 0 at 1: $p(1) = 0 $

so now i know that the solution will be something like:

$y_p = x^2(Ax + B)e^x$
where the parentheses show that it is a linear operator on the last coefficient.

I am told to use the exponential shift rule on $e^x$ so i believe that it is this:

$y_p = x^2e^x(Ax + B + 1)$

but i'm not sure if this is correct or where to go from here. because now the operator is operating on a coefficient of 1?

thanks for your help

linear-algebra differential-equations stability-in-odes

share|cite|improve this question

edited Jul 16 at 23:31

Key Flex

4,396425

asked Jul 16 at 23:27

entercaspa

334

add a comment | 

up vote
1
down vote

favorite

I am trying to find the particular solution to the $y^4 -2y'' +y = xe^x $
and currently am misunderstanding what to do.

My steps:

the polynomial operator concerned is $p(s)= s^4 -2s^2 + 1 $
which 0 at 1: $p(1) = 0 $

so now i know that the solution will be something like:

$y_p = x^2(Ax + B)e^x$
where the parentheses show that it is a linear operator on the last coefficient.

I am told to use the exponential shift rule on $e^x$ so i believe that it is this:

$y_p = x^2e^x(Ax + B + 1)$

but i'm not sure if this is correct or where to go from here. because now the operator is operating on a coefficient of 1?

thanks for your help

linear-algebra differential-equations stability-in-odes

share|cite|improve this question

edited Jul 16 at 23:31

Key Flex

4,396425

asked Jul 16 at 23:27

entercaspa

334

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I am trying to find the particular solution to the $y^4 -2y'' +y = xe^x $
and currently am misunderstanding what to do.

My steps:

the polynomial operator concerned is $p(s)= s^4 -2s^2 + 1 $
which 0 at 1: $p(1) = 0 $

so now i know that the solution will be something like:

$y_p = x^2(Ax + B)e^x$
where the parentheses show that it is a linear operator on the last coefficient.

I am told to use the exponential shift rule on $e^x$ so i believe that it is this:

$y_p = x^2e^x(Ax + B + 1)$

but i'm not sure if this is correct or where to go from here. because now the operator is operating on a coefficient of 1?

thanks for your help

linear-algebra differential-equations stability-in-odes

share|cite|improve this question

edited Jul 16 at 23:31

Key Flex

4,396425

asked Jul 16 at 23:27

entercaspa

334

I am trying to find the particular solution to the $y^4 -2y'' +y = xe^x $
and currently am misunderstanding what to do.

My steps:

the polynomial operator concerned is $p(s)= s^4 -2s^2 + 1 $
which 0 at 1: $p(1) = 0 $

so now i know that the solution will be something like:

$y_p = x^2(Ax + B)e^x$
where the parentheses show that it is a linear operator on the last coefficient.

I am told to use the exponential shift rule on $e^x$ so i believe that it is this:

$y_p = x^2e^x(Ax + B + 1)$

but i'm not sure if this is correct or where to go from here. because now the operator is operating on a coefficient of 1?

thanks for your help

linear-algebra differential-equations stability-in-odes

share|cite|improve this question

edited Jul 16 at 23:31

Key Flex

4,396425

asked Jul 16 at 23:27

entercaspa

334

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 23:31

Key Flex

4,396425

edited Jul 16 at 23:31

Key Flex

4,396425

edited Jul 16 at 23:31

Key Flex

4,396425

4,396425

asked Jul 16 at 23:27

entercaspa

334

asked Jul 16 at 23:27

entercaspa

334

asked Jul 16 at 23:27

entercaspa

334

334

add a comment | 

add a comment | 

1 Answer
1

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oldest

votes

up vote
2
down vote

$$y^4 -2y'' +y = xe^x$$
The caracteristic polynomial is
$$implies r^4-2r^2+1=0$$
$$(r^2-1)^2=0 $$
$$(r-1)^2(r+1)^2=0 implies r=1,-1$$
The solution to the homogeneous equation is
$$y_h=c_1e^x+c_2xe^x+c_3e^-x+c_4xe^-x$$
for the particular solution try
$$y_p=(Ax^3+Bx^2)e^x$$

---

Another method

$$y^4 -2y'' +y = xe^x$$
$$y^4 -y''-y'' +y = xe^x$$
$$(y'' -y)''-(y'' -y) = xe^x$$
Substitute $z=y''-y$
$$z''-z=xe^x$$
it's linear of second order

share|cite|improve this answer

edited Jul 17 at 15:05

answered Jul 16 at 23:38

Isham

10.6k3829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am slightly confused because the book says try $ Y_p=x^2(Ax+B)e^x$
  – entercaspa
  Jul 16 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  its the same as the particular solution I wrote @entercaspa $$x^2(Ax+B)e^x=(Ax^3+Bx^2)e^x$$
  – Isham
  Jul 16 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i see that now, i did not know that was a rule of linear operators
  – entercaspa
  Jul 16 at 23:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  differentiate and plug that in the original equation to get the value of A and B @entercaspa
  – Isham
  Jul 16 at 23:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so is that the same as $y_p = x^2e^x(Ax+B+1)$ following the exponential shift rule?
  – entercaspa
  Jul 16 at 23:44
  

  
  
  
  

 | 
show 3 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

$$y^4 -2y'' +y = xe^x$$
The caracteristic polynomial is
$$implies r^4-2r^2+1=0$$
$$(r^2-1)^2=0 $$
$$(r-1)^2(r+1)^2=0 implies r=1,-1$$
The solution to the homogeneous equation is
$$y_h=c_1e^x+c_2xe^x+c_3e^-x+c_4xe^-x$$
for the particular solution try
$$y_p=(Ax^3+Bx^2)e^x$$

---

Another method

$$y^4 -2y'' +y = xe^x$$
$$y^4 -y''-y'' +y = xe^x$$
$$(y'' -y)''-(y'' -y) = xe^x$$
Substitute $z=y''-y$
$$z''-z=xe^x$$
it's linear of second order

share|cite|improve this answer

edited Jul 17 at 15:05

answered Jul 16 at 23:38

Isham

10.6k3829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am slightly confused because the book says try $ Y_p=x^2(Ax+B)e^x$
  – entercaspa
  Jul 16 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  its the same as the particular solution I wrote @entercaspa $$x^2(Ax+B)e^x=(Ax^3+Bx^2)e^x$$
  – Isham
  Jul 16 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i see that now, i did not know that was a rule of linear operators
  – entercaspa
  Jul 16 at 23:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  differentiate and plug that in the original equation to get the value of A and B @entercaspa
  – Isham
  Jul 16 at 23:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so is that the same as $y_p = x^2e^x(Ax+B+1)$ following the exponential shift rule?
  – entercaspa
  Jul 16 at 23:44
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote

$$y^4 -2y'' +y = xe^x$$
The caracteristic polynomial is
$$implies r^4-2r^2+1=0$$
$$(r^2-1)^2=0 $$
$$(r-1)^2(r+1)^2=0 implies r=1,-1$$
The solution to the homogeneous equation is
$$y_h=c_1e^x+c_2xe^x+c_3e^-x+c_4xe^-x$$
for the particular solution try
$$y_p=(Ax^3+Bx^2)e^x$$

---

Another method

$$y^4 -2y'' +y = xe^x$$
$$y^4 -y''-y'' +y = xe^x$$
$$(y'' -y)''-(y'' -y) = xe^x$$
Substitute $z=y''-y$
$$z''-z=xe^x$$
it's linear of second order

share|cite|improve this answer

edited Jul 17 at 15:05

answered Jul 16 at 23:38

Isham

10.6k3829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am slightly confused because the book says try $ Y_p=x^2(Ax+B)e^x$
  – entercaspa
  Jul 16 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  its the same as the particular solution I wrote @entercaspa $$x^2(Ax+B)e^x=(Ax^3+Bx^2)e^x$$
  – Isham
  Jul 16 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i see that now, i did not know that was a rule of linear operators
  – entercaspa
  Jul 16 at 23:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  differentiate and plug that in the original equation to get the value of A and B @entercaspa
  – Isham
  Jul 16 at 23:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so is that the same as $y_p = x^2e^x(Ax+B+1)$ following the exponential shift rule?
  – entercaspa
  Jul 16 at 23:44
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote

up vote
2
down vote

$$y^4 -2y'' +y = xe^x$$
The caracteristic polynomial is
$$implies r^4-2r^2+1=0$$
$$(r^2-1)^2=0 $$
$$(r-1)^2(r+1)^2=0 implies r=1,-1$$
The solution to the homogeneous equation is
$$y_h=c_1e^x+c_2xe^x+c_3e^-x+c_4xe^-x$$
for the particular solution try
$$y_p=(Ax^3+Bx^2)e^x$$

---

Another method

$$y^4 -2y'' +y = xe^x$$
$$y^4 -y''-y'' +y = xe^x$$
$$(y'' -y)''-(y'' -y) = xe^x$$
Substitute $z=y''-y$
$$z''-z=xe^x$$
it's linear of second order

share|cite|improve this answer

edited Jul 17 at 15:05

answered Jul 16 at 23:38

Isham

10.6k3829

$$y^4 -2y'' +y = xe^x$$
The caracteristic polynomial is
$$implies r^4-2r^2+1=0$$
$$(r^2-1)^2=0 $$
$$(r-1)^2(r+1)^2=0 implies r=1,-1$$
The solution to the homogeneous equation is
$$y_h=c_1e^x+c_2xe^x+c_3e^-x+c_4xe^-x$$
for the particular solution try
$$y_p=(Ax^3+Bx^2)e^x$$

---

Another method

$$y^4 -2y'' +y = xe^x$$
$$y^4 -y''-y'' +y = xe^x$$
$$(y'' -y)''-(y'' -y) = xe^x$$
Substitute $z=y''-y$
$$z''-z=xe^x$$
it's linear of second order

share|cite|improve this answer

edited Jul 17 at 15:05

answered Jul 16 at 23:38

Isham

10.6k3829

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 15:05

edited Jul 17 at 15:05

edited Jul 17 at 15:05

answered Jul 16 at 23:38

Isham

10.6k3829

answered Jul 16 at 23:38

Isham

10.6k3829

answered Jul 16 at 23:38

Isham

10.6k3829

10.6k3829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am slightly confused because the book says try $ Y_p=x^2(Ax+B)e^x$
  – entercaspa
  Jul 16 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  its the same as the particular solution I wrote @entercaspa $$x^2(Ax+B)e^x=(Ax^3+Bx^2)e^x$$
  – Isham
  Jul 16 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i see that now, i did not know that was a rule of linear operators
  – entercaspa
  Jul 16 at 23:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  differentiate and plug that in the original equation to get the value of A and B @entercaspa
  – Isham
  Jul 16 at 23:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so is that the same as $y_p = x^2e^x(Ax+B+1)$ following the exponential shift rule?
  – entercaspa
  Jul 16 at 23:44
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am slightly confused because the book says try $ Y_p=x^2(Ax+B)e^x$
  – entercaspa
  Jul 16 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  its the same as the particular solution I wrote @entercaspa $$x^2(Ax+B)e^x=(Ax^3+Bx^2)e^x$$
  – Isham
  Jul 16 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i see that now, i did not know that was a rule of linear operators
  – entercaspa
  Jul 16 at 23:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  differentiate and plug that in the original equation to get the value of A and B @entercaspa
  – Isham
  Jul 16 at 23:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so is that the same as $y_p = x^2e^x(Ax+B+1)$ following the exponential shift rule?
  – entercaspa
  Jul 16 at 23:44
  

  
  
  
  

I am slightly confused because the book says try $ Y_p=x^2(Ax+B)e^x$
– entercaspa
Jul 16 at 23:42

I am slightly confused because the book says try $ Y_p=x^2(Ax+B)e^x$
– entercaspa
Jul 16 at 23:42

its the same as the particular solution I wrote @entercaspa $$x^2(Ax+B)e^x=(Ax^3+Bx^2)e^x$$
– Isham
Jul 16 at 23:42

its the same as the particular solution I wrote @entercaspa $$x^2(Ax+B)e^x=(Ax^3+Bx^2)e^x$$
– Isham
Jul 16 at 23:42

i see that now, i did not know that was a rule of linear operators
– entercaspa
Jul 16 at 23:43

i see that now, i did not know that was a rule of linear operators
– entercaspa
Jul 16 at 23:43

differentiate and plug that in the original equation to get the value of A and B @entercaspa
– Isham
Jul 16 at 23:44

differentiate and plug that in the original equation to get the value of A and B @entercaspa
– Isham
Jul 16 at 23:44

so is that the same as $y_p = x^2e^x(Ax+B+1)$ following the exponential shift rule?
– entercaspa
Jul 16 at 23:44

so is that the same as $y_p = x^2e^x(Ax+B+1)$ following the exponential shift rule?
– entercaspa
Jul 16 at 23:44

 | 
show 3 more comments

 

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