Solve $t^2y''-2ty'+2y=6t^2+4ln(t)$

0

$begingroup$

I know that I should find a particular solution, but how ?

I used change of variable method, $t=e^u$ but it didn't work

ordinary-differential-equations

share|cite|improve this question

asked Jan 31 at 3:14

Pedro AlvarèsPedro Alvarès

946

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an Euler equation, it's characteristic polynomial is $r(r-1) -2r+2 =0$
  $endgroup$
  – LordVader007
  Jan 31 at 3:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See :tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx
  $endgroup$
  – LordVader007
  Jan 31 at 3:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Okay, but we should solve for the homogeneous form of the equation first right ?
  $endgroup$
  – Pedro Alvarès
  Jan 31 at 3:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes... once you find a fundamental set of solutions, you can use something like Variation of Parameters to solve for the inhomogeneous case.
  $endgroup$
  – LordVader007
  Jan 31 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The fundamental set of solutions should be: $phi_1(t) = t$ and $phi_2(t) = t^2$
  $endgroup$
  – LordVader007
  Jan 31 at 3:27
  
  
  

  
  
  
  

 | 
show 6 more comments

0

$begingroup$

I know that I should find a particular solution, but how ?

I used change of variable method, $t=e^u$ but it didn't work

ordinary-differential-equations

share|cite|improve this question

asked Jan 31 at 3:14

Pedro AlvarèsPedro Alvarès

946

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an Euler equation, it's characteristic polynomial is $r(r-1) -2r+2 =0$
  $endgroup$
  – LordVader007
  Jan 31 at 3:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See :tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx
  $endgroup$
  – LordVader007
  Jan 31 at 3:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Okay, but we should solve for the homogeneous form of the equation first right ?
  $endgroup$
  – Pedro Alvarès
  Jan 31 at 3:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes... once you find a fundamental set of solutions, you can use something like Variation of Parameters to solve for the inhomogeneous case.
  $endgroup$
  – LordVader007
  Jan 31 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The fundamental set of solutions should be: $phi_1(t) = t$ and $phi_2(t) = t^2$
  $endgroup$
  – LordVader007
  Jan 31 at 3:27
  
  
  

  
  
  
  

 | 
show 6 more comments

0

0

0

$begingroup$

I know that I should find a particular solution, but how ?

I used change of variable method, $t=e^u$ but it didn't work

ordinary-differential-equations

share|cite|improve this question

asked Jan 31 at 3:14

Pedro AlvarèsPedro Alvarès

946

$endgroup$

I know that I should find a particular solution, but how ?

I used change of variable method, $t=e^u$ but it didn't work

ordinary-differential-equations

ordinary-differential-equations

share|cite|improve this question

asked Jan 31 at 3:14

Pedro AlvarèsPedro Alvarès

946

share|cite|improve this question

asked Jan 31 at 3:14

Pedro AlvarèsPedro Alvarès

946

share|cite|improve this question

share|cite|improve this question

asked Jan 31 at 3:14

Pedro AlvarèsPedro Alvarès

946

asked Jan 31 at 3:14

Pedro AlvarèsPedro Alvarès

946

asked Jan 31 at 3:14

Pedro AlvarèsPedro Alvarès

946

946

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an Euler equation, it's characteristic polynomial is $r(r-1) -2r+2 =0$
  $endgroup$
  – LordVader007
  Jan 31 at 3:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See :tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx
  $endgroup$
  – LordVader007
  Jan 31 at 3:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Okay, but we should solve for the homogeneous form of the equation first right ?
  $endgroup$
  – Pedro Alvarès
  Jan 31 at 3:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes... once you find a fundamental set of solutions, you can use something like Variation of Parameters to solve for the inhomogeneous case.
  $endgroup$
  – LordVader007
  Jan 31 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The fundamental set of solutions should be: $phi_1(t) = t$ and $phi_2(t) = t^2$
  $endgroup$
  – LordVader007
  Jan 31 at 3:27
  
  
  

  
  
  
  

 | 
show 6 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an Euler equation, it's characteristic polynomial is $r(r-1) -2r+2 =0$
  $endgroup$
  – LordVader007
  Jan 31 at 3:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See :tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx
  $endgroup$
  – LordVader007
  Jan 31 at 3:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Okay, but we should solve for the homogeneous form of the equation first right ?
  $endgroup$
  – Pedro Alvarès
  Jan 31 at 3:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes... once you find a fundamental set of solutions, you can use something like Variation of Parameters to solve for the inhomogeneous case.
  $endgroup$
  – LordVader007
  Jan 31 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The fundamental set of solutions should be: $phi_1(t) = t$ and $phi_2(t) = t^2$
  $endgroup$
  – LordVader007
  Jan 31 at 3:27
  
  
  

  
  
  
  

$begingroup$
This is an Euler equation, it's characteristic polynomial is $r(r-1) -2r+2 =0$
$endgroup$
– LordVader007
Jan 31 at 3:15

$begingroup$
This is an Euler equation, it's characteristic polynomial is $r(r-1) -2r+2 =0$
$endgroup$
– LordVader007
Jan 31 at 3:15

$begingroup$
See :tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx
$endgroup$
– LordVader007
Jan 31 at 3:16

$begingroup$
See :tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx
$endgroup$
– LordVader007
Jan 31 at 3:16

$begingroup$
Okay, but we should solve for the homogeneous form of the equation first right ?
$endgroup$
– Pedro Alvarès
Jan 31 at 3:20

$begingroup$
Okay, but we should solve for the homogeneous form of the equation first right ?
$endgroup$
– Pedro Alvarès
Jan 31 at 3:20

$begingroup$
Yes... once you find a fundamental set of solutions, you can use something like Variation of Parameters to solve for the inhomogeneous case.
$endgroup$
– LordVader007
Jan 31 at 3:22

$begingroup$
Yes... once you find a fundamental set of solutions, you can use something like Variation of Parameters to solve for the inhomogeneous case.
$endgroup$
– LordVader007
Jan 31 at 3:22

$begingroup$
The fundamental set of solutions should be: $phi_1(t) = t$ and $phi_2(t) = t^2$
$endgroup$
– LordVader007
Jan 31 at 3:27

$begingroup$
The fundamental set of solutions should be: $phi_1(t) = t$ and $phi_2(t) = t^2$
$endgroup$
– LordVader007
Jan 31 at 3:27

 | 
show 6 more comments

1 Answer
1

active

oldest

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2

$begingroup$

Solve $$x^2y''-2xy'+2y=6x^2+4ln x.$$

We use change of variable method
$$x=e^t,quad t=ln x,quad frac{dt}{dx}=frac1x=e^{-t},$$
$$y'=frac{dy}{dx}=frac{dy}{dt}frac{dt}{dx}=y_te^{-t},$$
$$y''=frac{dy'}{dx}=frac{dy'}{dt}frac{dt}{dx}=frac{d(y_te^{-t})}{dt}e^{-t}=(y''_{tt}-y'_t)e^{-2t}$$
Then
$$xy'=y'_t,quad x^2y''=y''_{tt}-y'_t.$$
We get linear dif. equation with constants coefficients
$$y''_{tt}-3y'_t+2y=6e^{2t}+4t,$$
$$y=y_h+y_p,$$
$$y_h=c_1e^t+c_2e^{2t}$$
With method of undetermined coefficients we find particular solution
$$y_p=Ate^{2t}+Bt+C,\y_p=6te^{2t}+2t+3.$$
Then general solution is
$$y=c_1e^t+c_2e^{2t}+6te^{2t}+2t+3\
=c_1x+c_2x^2+6x^2ln x+2ln x +3.
$$

share|cite|improve this answer

answered Jan 31 at 9:46

Aleksas DomarkasAleksas Domarkas

1,62317

$endgroup$

add a comment | 

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1 Answer
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active

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1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

Solve $$x^2y''-2xy'+2y=6x^2+4ln x.$$

We use change of variable method
$$x=e^t,quad t=ln x,quad frac{dt}{dx}=frac1x=e^{-t},$$
$$y'=frac{dy}{dx}=frac{dy}{dt}frac{dt}{dx}=y_te^{-t},$$
$$y''=frac{dy'}{dx}=frac{dy'}{dt}frac{dt}{dx}=frac{d(y_te^{-t})}{dt}e^{-t}=(y''_{tt}-y'_t)e^{-2t}$$
Then
$$xy'=y'_t,quad x^2y''=y''_{tt}-y'_t.$$
We get linear dif. equation with constants coefficients
$$y''_{tt}-3y'_t+2y=6e^{2t}+4t,$$
$$y=y_h+y_p,$$
$$y_h=c_1e^t+c_2e^{2t}$$
With method of undetermined coefficients we find particular solution
$$y_p=Ate^{2t}+Bt+C,\y_p=6te^{2t}+2t+3.$$
Then general solution is
$$y=c_1e^t+c_2e^{2t}+6te^{2t}+2t+3\
=c_1x+c_2x^2+6x^2ln x+2ln x +3.
$$

share|cite|improve this answer

answered Jan 31 at 9:46

Aleksas DomarkasAleksas Domarkas

1,62317

$endgroup$

add a comment | 

2

$begingroup$

Solve $$x^2y''-2xy'+2y=6x^2+4ln x.$$

We use change of variable method
$$x=e^t,quad t=ln x,quad frac{dt}{dx}=frac1x=e^{-t},$$
$$y'=frac{dy}{dx}=frac{dy}{dt}frac{dt}{dx}=y_te^{-t},$$
$$y''=frac{dy'}{dx}=frac{dy'}{dt}frac{dt}{dx}=frac{d(y_te^{-t})}{dt}e^{-t}=(y''_{tt}-y'_t)e^{-2t}$$
Then
$$xy'=y'_t,quad x^2y''=y''_{tt}-y'_t.$$
We get linear dif. equation with constants coefficients
$$y''_{tt}-3y'_t+2y=6e^{2t}+4t,$$
$$y=y_h+y_p,$$
$$y_h=c_1e^t+c_2e^{2t}$$
With method of undetermined coefficients we find particular solution
$$y_p=Ate^{2t}+Bt+C,\y_p=6te^{2t}+2t+3.$$
Then general solution is
$$y=c_1e^t+c_2e^{2t}+6te^{2t}+2t+3\
=c_1x+c_2x^2+6x^2ln x+2ln x +3.
$$

share|cite|improve this answer

answered Jan 31 at 9:46

Aleksas DomarkasAleksas Domarkas

1,62317

$endgroup$

add a comment | 

2

2

2

$begingroup$

Solve $$x^2y''-2xy'+2y=6x^2+4ln x.$$

We use change of variable method
$$x=e^t,quad t=ln x,quad frac{dt}{dx}=frac1x=e^{-t},$$
$$y'=frac{dy}{dx}=frac{dy}{dt}frac{dt}{dx}=y_te^{-t},$$
$$y''=frac{dy'}{dx}=frac{dy'}{dt}frac{dt}{dx}=frac{d(y_te^{-t})}{dt}e^{-t}=(y''_{tt}-y'_t)e^{-2t}$$
Then
$$xy'=y'_t,quad x^2y''=y''_{tt}-y'_t.$$
We get linear dif. equation with constants coefficients
$$y''_{tt}-3y'_t+2y=6e^{2t}+4t,$$
$$y=y_h+y_p,$$
$$y_h=c_1e^t+c_2e^{2t}$$
With method of undetermined coefficients we find particular solution
$$y_p=Ate^{2t}+Bt+C,\y_p=6te^{2t}+2t+3.$$
Then general solution is
$$y=c_1e^t+c_2e^{2t}+6te^{2t}+2t+3\
=c_1x+c_2x^2+6x^2ln x+2ln x +3.
$$

share|cite|improve this answer

answered Jan 31 at 9:46

Aleksas DomarkasAleksas Domarkas

1,62317

$endgroup$

Solve $$x^2y''-2xy'+2y=6x^2+4ln x.$$

We use change of variable method
$$x=e^t,quad t=ln x,quad frac{dt}{dx}=frac1x=e^{-t},$$
$$y'=frac{dy}{dx}=frac{dy}{dt}frac{dt}{dx}=y_te^{-t},$$
$$y''=frac{dy'}{dx}=frac{dy'}{dt}frac{dt}{dx}=frac{d(y_te^{-t})}{dt}e^{-t}=(y''_{tt}-y'_t)e^{-2t}$$
Then
$$xy'=y'_t,quad x^2y''=y''_{tt}-y'_t.$$
We get linear dif. equation with constants coefficients
$$y''_{tt}-3y'_t+2y=6e^{2t}+4t,$$
$$y=y_h+y_p,$$
$$y_h=c_1e^t+c_2e^{2t}$$
With method of undetermined coefficients we find particular solution
$$y_p=Ate^{2t}+Bt+C,\y_p=6te^{2t}+2t+3.$$
Then general solution is
$$y=c_1e^t+c_2e^{2t}+6te^{2t}+2t+3\
=c_1x+c_2x^2+6x^2ln x+2ln x +3.
$$

share|cite|improve this answer

answered Jan 31 at 9:46

Aleksas DomarkasAleksas Domarkas

1,62317

share|cite|improve this answer

share|cite|improve this answer

answered Jan 31 at 9:46

Aleksas DomarkasAleksas Domarkas

1,62317

answered Jan 31 at 9:46

Aleksas DomarkasAleksas Domarkas

1,62317

answered Jan 31 at 9:46

Aleksas DomarkasAleksas Domarkas

1,62317

1,62317

add a comment | 

add a comment | 

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