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Find all the constants $ainmathbbR$ so that the differential equations have solutions in common.

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2

1

$begingroup$

Find all the constants $ainmathbbR$ so that the differential equations:
$$(1):y''+ay'-2y=0$$
$$(2):y''-2y'+ay=0$$

have solutions in common.

I think I can get all solutions just by solving the system of the caracteristic polynomials for $lambda$:
$beginalign*
lambda^2+alambda-2 &=0\
lambda^2-2lambda+a &=0
endalign*$ $Rightarrow$ $beginalign*
lambda &=dfrac-apmsqrta^2+42\
lambda &=dfrac2pm2sqrt1-a2
endalign*$


But then I have to solve for all the case for when the $pm$ changes so that I have 4 cases $(+,+);(+,-);(-,+);(-,-)$

This is pretty slow so I was wondering if there was an smarter way to solve for a.

ordinary-differential-equations

share|cite|improve this question

asked Mar 13 at 18:44

Sebastian CorSebastian Cor

12610

$endgroup$

add a comment | 

2

1

$begingroup$

Find all the constants $ainmathbbR$ so that the differential equations:
$$(1):y''+ay'-2y=0$$
$$(2):y''-2y'+ay=0$$

have solutions in common.

I think I can get all solutions just by solving the system of the caracteristic polynomials for $lambda$:
$beginalign*
lambda^2+alambda-2 &=0\
lambda^2-2lambda+a &=0
endalign*$ $Rightarrow$ $beginalign*
lambda &=dfrac-apmsqrta^2+42\
lambda &=dfrac2pm2sqrt1-a2
endalign*$


But then I have to solve for all the case for when the $pm$ changes so that I have 4 cases $(+,+);(+,-);(-,+);(-,-)$

This is pretty slow so I was wondering if there was an smarter way to solve for a.

ordinary-differential-equations

share|cite|improve this question

asked Mar 13 at 18:44

Sebastian CorSebastian Cor

12610

$endgroup$

add a comment | 

$begingroup$

Find all the constants $ainmathbbR$ so that the differential equations:
$$(1):y''+ay'-2y=0$$
$$(2):y''-2y'+ay=0$$

have solutions in common.

I think I can get all solutions just by solving the system of the caracteristic polynomials for $lambda$:
$beginalign*
lambda^2+alambda-2 &=0\
lambda^2-2lambda+a &=0
endalign*$ $Rightarrow$ $beginalign*
lambda &=dfrac-apmsqrta^2+42\
lambda &=dfrac2pm2sqrt1-a2
endalign*$


But then I have to solve for all the case for when the $pm$ changes so that I have 4 cases $(+,+);(+,-);(-,+);(-,-)$

This is pretty slow so I was wondering if there was an smarter way to solve for a.

ordinary-differential-equations

share|cite|improve this question

asked Mar 13 at 18:44

Sebastian CorSebastian Cor

12610

$endgroup$

share|cite|improve this question

asked Mar 13 at 18:44

Sebastian CorSebastian Cor

12610

share|cite|improve this question

asked Mar 13 at 18:44

Sebastian CorSebastian Cor

12610

asked Mar 13 at 18:44

Sebastian CorSebastian Cor

12610

asked Mar 13 at 18:44

Sebastian CorSebastian Cor

12610

1 Answer
1

active

oldest

votes

1

$begingroup$

I assume that what you mean is non zero solutions (because $y=0$ is solution to both differential equations, whatever the value of a).

If $y$ is solution it is also solution of the differential equation obtained by taking the difference of these two equations, i.e.,

$$(a+2)y'=(a+2)y tag1$$

Case 1 : if $a neq -2$ : one can conclude from (1) that $y$ is such that $y'=y$, i.e.

$$y=Ke^t textwith K neq 0.tag2$$

Plugging (2) for example in the first equation, one gets

$$Ke^t+aKe^t-2Ke^t=0$$

meaning that $1+a-2=0$, thus solution (2) is possible iff $a=1$.

Case 2 : if, on the contrary, $a=-2$, both initial equations give :

$$y''-2y'-2y=0,$$ and this unique non parametric equation has clearly solutions (that aren't asked).

Thus the answer is :

There are two values $a=1$ and $a=-2$ satisfying the condition.

share|cite|improve this answer

edited Mar 13 at 19:58

answered Mar 13 at 18:56

Jean MarieJean Marie

30.8k42154

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn't $a=-2$ be a solution as well because then the two differential equations are the same and have solutions in common?
  $endgroup$
  – Paras Khosla
  Mar 13 at 19:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Lutzl and Paras Khosla : you are right. Corrected...
  $endgroup$
  – Jean Marie
  Mar 13 at 19:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is exactly what I was looking for! Thanks.
  $endgroup$
  – Sebastian Cor
  Mar 13 at 20:04
  

  
  
  
  

add a comment | 

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1

$begingroup$

I assume that what you mean is non zero solutions (because $y=0$ is solution to both differential equations, whatever the value of a).

If $y$ is solution it is also solution of the differential equation obtained by taking the difference of these two equations, i.e.,

$$(a+2)y'=(a+2)y tag1$$

Case 1 : if $a neq -2$ : one can conclude from (1) that $y$ is such that $y'=y$, i.e.

$$y=Ke^t textwith K neq 0.tag2$$

Plugging (2) for example in the first equation, one gets

$$Ke^t+aKe^t-2Ke^t=0$$

meaning that $1+a-2=0$, thus solution (2) is possible iff $a=1$.

Case 2 : if, on the contrary, $a=-2$, both initial equations give :

$$y''-2y'-2y=0,$$ and this unique non parametric equation has clearly solutions (that aren't asked).

Thus the answer is :

There are two values $a=1$ and $a=-2$ satisfying the condition.

share|cite|improve this answer

edited Mar 13 at 19:58

answered Mar 13 at 18:56

Jean MarieJean Marie

30.8k42154

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn't $a=-2$ be a solution as well because then the two differential equations are the same and have solutions in common?
  $endgroup$
  – Paras Khosla
  Mar 13 at 19:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Lutzl and Paras Khosla : you are right. Corrected...
  $endgroup$
  – Jean Marie
  Mar 13 at 19:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is exactly what I was looking for! Thanks.
  $endgroup$
  – Sebastian Cor
  Mar 13 at 20:04
  

  
  
  
  

add a comment | 

1

$begingroup$

I assume that what you mean is non zero solutions (because $y=0$ is solution to both differential equations, whatever the value of a).

If $y$ is solution it is also solution of the differential equation obtained by taking the difference of these two equations, i.e.,

$$(a+2)y'=(a+2)y tag1$$

Case 1 : if $a neq -2$ : one can conclude from (1) that $y$ is such that $y'=y$, i.e.

$$y=Ke^t textwith K neq 0.tag2$$

Plugging (2) for example in the first equation, one gets

$$Ke^t+aKe^t-2Ke^t=0$$

meaning that $1+a-2=0$, thus solution (2) is possible iff $a=1$.

Case 2 : if, on the contrary, $a=-2$, both initial equations give :

$$y''-2y'-2y=0,$$ and this unique non parametric equation has clearly solutions (that aren't asked).

Thus the answer is :

There are two values $a=1$ and $a=-2$ satisfying the condition.

share|cite|improve this answer

edited Mar 13 at 19:58

answered Mar 13 at 18:56

Jean MarieJean Marie

30.8k42154

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn't $a=-2$ be a solution as well because then the two differential equations are the same and have solutions in common?
  $endgroup$
  – Paras Khosla
  Mar 13 at 19:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Lutzl and Paras Khosla : you are right. Corrected...
  $endgroup$
  – Jean Marie
  Mar 13 at 19:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is exactly what I was looking for! Thanks.
  $endgroup$
  – Sebastian Cor
  Mar 13 at 20:04
  

  
  
  
  

add a comment | 

$begingroup$

I assume that what you mean is non zero solutions (because $y=0$ is solution to both differential equations, whatever the value of a).

If $y$ is solution it is also solution of the differential equation obtained by taking the difference of these two equations, i.e.,

$$(a+2)y'=(a+2)y tag1$$

Case 1 : if $a neq -2$ : one can conclude from (1) that $y$ is such that $y'=y$, i.e.

$$y=Ke^t textwith K neq 0.tag2$$

Plugging (2) for example in the first equation, one gets

$$Ke^t+aKe^t-2Ke^t=0$$

meaning that $1+a-2=0$, thus solution (2) is possible iff $a=1$.

Case 2 : if, on the contrary, $a=-2$, both initial equations give :

$$y''-2y'-2y=0,$$ and this unique non parametric equation has clearly solutions (that aren't asked).

Thus the answer is :

There are two values $a=1$ and $a=-2$ satisfying the condition.

share|cite|improve this answer

edited Mar 13 at 19:58

answered Mar 13 at 18:56

Jean MarieJean Marie

30.8k42154

$endgroup$

share|cite|improve this answer

edited Mar 13 at 19:58

answered Mar 13 at 18:56

Jean MarieJean Marie

30.8k42154

answered Mar 13 at 18:56

Jean MarieJean Marie

30.8k42154

answered Mar 13 at 18:56

Jean MarieJean Marie

30.8k42154