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Particular answer to a differential Equation Involving Delta function and Heaviside without Laplace

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0

$begingroup$

Please help me find the particular answer for ODE's Like this:

$$4y''+5y'+3y+2=8h''+7h'+6h$$

$h(x)=$Heaviside function

ordinary-differential-equations

share|cite|improve this question

edited Mar 13 at 17:05

dantopa

6,63342245

asked Mar 13 at 16:42

Ali NaderinejadAli Naderinejad

1

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
  $endgroup$
  – dantopa
  Mar 13 at 17:04
  

  
  
  
  

add a comment | 

0

$begingroup$

Please help me find the particular answer for ODE's Like this:

$$4y''+5y'+3y+2=8h''+7h'+6h$$

$h(x)=$Heaviside function

ordinary-differential-equations

share|cite|improve this question

edited Mar 13 at 17:05

dantopa

6,63342245

asked Mar 13 at 16:42

Ali NaderinejadAli Naderinejad

1

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
  $endgroup$
  – dantopa
  Mar 13 at 17:04
  

  
  
  
  

add a comment | 

$begingroup$

Please help me find the particular answer for ODE's Like this:

$$4y''+5y'+3y+2=8h''+7h'+6h$$

$h(x)=$Heaviside function

ordinary-differential-equations

share|cite|improve this question

edited Mar 13 at 17:05

dantopa

6,63342245

asked Mar 13 at 16:42

Ali NaderinejadAli Naderinejad

1

$endgroup$

share|cite|improve this question

edited Mar 13 at 17:05

dantopa

6,63342245

asked Mar 13 at 16:42

Ali NaderinejadAli Naderinejad

1

share|cite|improve this question

edited Mar 13 at 17:05

dantopa

6,63342245

asked Mar 13 at 16:42

Ali NaderinejadAli Naderinejad

1

edited Mar 13 at 17:05

dantopa

6,63342245

edited Mar 13 at 17:05

dantopa

6,63342245

asked Mar 13 at 16:42

Ali NaderinejadAli Naderinejad

1

asked Mar 13 at 16:42

Ali NaderinejadAli Naderinejad

1

2 Answers
2

active

oldest

votes

1

$begingroup$

Set $u=y-2h$, then this new function will have the equation
$$
(4u''+8h'')+(5u'+10h')+(3u+6h)+2=8h''+7h'+6h
\iff\
4u''+5u'+3u+2=-3h'
$$
In a next step you can also absorb the first derivative in $u$ by adding an anti-derivative of $h$. Let $H(x)=max(0,x)$ and set $v=u+frac34H$, then
$$
(4v''-3h')+(5v'-frac{15}4h)+(3v-frac94H)+2=-3h'
\iff\
4v''+5v'+3v+2=frac{15}4h+frac94H
$$
You could continue to absorb the singularities or directly solve this ODE with piecewise continuous right side.

share|cite|improve this answer

answered Mar 13 at 18:10

LutzLLutzL

59.8k42057

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3147624/… can you help me with this question? thanks!
  $endgroup$
  – Mikey Spivak
  Mar 15 at 1:11
  

  
  
  
  

add a comment | 

0

$begingroup$

We can solve an equation like this by solving the equation in "well-behaved" regions of the real line and then integrating the ODE in a region around the singular points to find matching boundary conditions. I'll sketch out how this is done here and leave the details as an exercise for the reader.

Consider the equation
$$
4y''+5y'+3y+2=8h''+7h'+6h
$$
The right-hand side of this equation is equal to $6$ when $x > 0$, and equal to $0$ when $x < 0$. In these regions, we can therefore write
$$
4 y'' + 5 y' + 3y = begin{cases} 4 & x > 0 \ -2 & x < 0 end{cases}
$$
The solutions to this equation will then be
$$
y(t) = begin{cases} A_+ e^{alpha_1 x} + B_+ e^{alpha_2 x} + frac{4}{3} & x> 0 \ A_- e^{alpha_1 x} + B_- e^{alpha_2 x} - frac{2}{3} & x> 0
end{cases},
$$
where $alpha_1$ and $alpha_2$ are the roots of the characteristic polynomial $4 alpha^2 + 5 alpha + 3 = 0$, and $A_pm$ and $B_pm$ are as-yet undetermined coefficients.

We now integrate the original ODE over the region $x = [-epsilon, epsilon]$:
$$
4 left[ y' right]_{-epsilon}^epsilon + 5 left[ y right]_{-epsilon}^epsilon + int_{-epsilon}^{epsilon} (3y + 2) , dx = 8 left[ h' right]_{-epsilon}^epsilon + 7 left[ h right]_{-epsilon}^epsilon + 6epsilon
$$
In the limit as $epsilon to 0$, this becomes
$$
4 left[ y' right]_{0_-}^{0_+} + 5 left[ y right]_{0_-}^{0_+} = 7.
$$
This leads to a condition on the unknown coefficients $A_pm$ and $B_pm$. Similarly, integrating the equation twice over the same region and taking the $epsilon to 0$ limit yields
$$
4 left[ y right]_{0_-}^{0_+} = 8.
$$

The two discontinuity equations above give us two equations for the three unknowns $A_pm$ and $B_pm$. The other two equations are determined by imposing appropriate conditions as $x to -infty$; for example, if it must approach a constant, then we must have $A_- = B_- = 0$.

share|cite|improve this answer

answered Mar 13 at 19:09

Michael SeifertMichael Seifert

5,062625

$endgroup$

add a comment | 

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1

$begingroup$

Set $u=y-2h$, then this new function will have the equation
$$
(4u''+8h'')+(5u'+10h')+(3u+6h)+2=8h''+7h'+6h
\iff\
4u''+5u'+3u+2=-3h'
$$
In a next step you can also absorb the first derivative in $u$ by adding an anti-derivative of $h$. Let $H(x)=max(0,x)$ and set $v=u+frac34H$, then
$$
(4v''-3h')+(5v'-frac{15}4h)+(3v-frac94H)+2=-3h'
\iff\
4v''+5v'+3v+2=frac{15}4h+frac94H
$$
You could continue to absorb the singularities or directly solve this ODE with piecewise continuous right side.

share|cite|improve this answer

answered Mar 13 at 18:10

LutzLLutzL

59.8k42057

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3147624/… can you help me with this question? thanks!
  $endgroup$
  – Mikey Spivak
  Mar 15 at 1:11
  

  
  
  
  

add a comment | 

1

$begingroup$

Set $u=y-2h$, then this new function will have the equation
$$
(4u''+8h'')+(5u'+10h')+(3u+6h)+2=8h''+7h'+6h
\iff\
4u''+5u'+3u+2=-3h'
$$
In a next step you can also absorb the first derivative in $u$ by adding an anti-derivative of $h$. Let $H(x)=max(0,x)$ and set $v=u+frac34H$, then
$$
(4v''-3h')+(5v'-frac{15}4h)+(3v-frac94H)+2=-3h'
\iff\
4v''+5v'+3v+2=frac{15}4h+frac94H
$$
You could continue to absorb the singularities or directly solve this ODE with piecewise continuous right side.

share|cite|improve this answer

answered Mar 13 at 18:10

LutzLLutzL

59.8k42057

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3147624/… can you help me with this question? thanks!
  $endgroup$
  – Mikey Spivak
  Mar 15 at 1:11
  

  
  
  
  

add a comment | 

$begingroup$

Set $u=y-2h$, then this new function will have the equation
$$
(4u''+8h'')+(5u'+10h')+(3u+6h)+2=8h''+7h'+6h
\iff\
4u''+5u'+3u+2=-3h'
$$
In a next step you can also absorb the first derivative in $u$ by adding an anti-derivative of $h$. Let $H(x)=max(0,x)$ and set $v=u+frac34H$, then
$$
(4v''-3h')+(5v'-frac{15}4h)+(3v-frac94H)+2=-3h'
\iff\
4v''+5v'+3v+2=frac{15}4h+frac94H
$$
You could continue to absorb the singularities or directly solve this ODE with piecewise continuous right side.

share|cite|improve this answer

answered Mar 13 at 18:10

LutzLLutzL

59.8k42057

$endgroup$

share|cite|improve this answer

answered Mar 13 at 18:10

LutzLLutzL

59.8k42057

answered Mar 13 at 18:10

LutzLLutzL

59.8k42057

answered Mar 13 at 18:10

LutzLLutzL

59.8k42057

0

$begingroup$

We can solve an equation like this by solving the equation in "well-behaved" regions of the real line and then integrating the ODE in a region around the singular points to find matching boundary conditions. I'll sketch out how this is done here and leave the details as an exercise for the reader.

Consider the equation
$$
4y''+5y'+3y+2=8h''+7h'+6h
$$
The right-hand side of this equation is equal to $6$ when $x > 0$, and equal to $0$ when $x < 0$. In these regions, we can therefore write
$$
4 y'' + 5 y' + 3y = begin{cases} 4 & x > 0 \ -2 & x < 0 end{cases}
$$
The solutions to this equation will then be
$$
y(t) = begin{cases} A_+ e^{alpha_1 x} + B_+ e^{alpha_2 x} + frac{4}{3} & x> 0 \ A_- e^{alpha_1 x} + B_- e^{alpha_2 x} - frac{2}{3} & x> 0
end{cases},
$$
where $alpha_1$ and $alpha_2$ are the roots of the characteristic polynomial $4 alpha^2 + 5 alpha + 3 = 0$, and $A_pm$ and $B_pm$ are as-yet undetermined coefficients.

We now integrate the original ODE over the region $x = [-epsilon, epsilon]$:
$$
4 left[ y' right]_{-epsilon}^epsilon + 5 left[ y right]_{-epsilon}^epsilon + int_{-epsilon}^{epsilon} (3y + 2) , dx = 8 left[ h' right]_{-epsilon}^epsilon + 7 left[ h right]_{-epsilon}^epsilon + 6epsilon
$$
In the limit as $epsilon to 0$, this becomes
$$
4 left[ y' right]_{0_-}^{0_+} + 5 left[ y right]_{0_-}^{0_+} = 7.
$$
This leads to a condition on the unknown coefficients $A_pm$ and $B_pm$. Similarly, integrating the equation twice over the same region and taking the $epsilon to 0$ limit yields
$$
4 left[ y right]_{0_-}^{0_+} = 8.
$$

The two discontinuity equations above give us two equations for the three unknowns $A_pm$ and $B_pm$. The other two equations are determined by imposing appropriate conditions as $x to -infty$; for example, if it must approach a constant, then we must have $A_- = B_- = 0$.

share|cite|improve this answer

answered Mar 13 at 19:09

Michael SeifertMichael Seifert

5,062625

$endgroup$

add a comment | 

0

$begingroup$

We can solve an equation like this by solving the equation in "well-behaved" regions of the real line and then integrating the ODE in a region around the singular points to find matching boundary conditions. I'll sketch out how this is done here and leave the details as an exercise for the reader.

Consider the equation
$$
4y''+5y'+3y+2=8h''+7h'+6h
$$
The right-hand side of this equation is equal to $6$ when $x > 0$, and equal to $0$ when $x < 0$. In these regions, we can therefore write
$$
4 y'' + 5 y' + 3y = begin{cases} 4 & x > 0 \ -2 & x < 0 end{cases}
$$
The solutions to this equation will then be
$$
y(t) = begin{cases} A_+ e^{alpha_1 x} + B_+ e^{alpha_2 x} + frac{4}{3} & x> 0 \ A_- e^{alpha_1 x} + B_- e^{alpha_2 x} - frac{2}{3} & x> 0
end{cases},
$$
where $alpha_1$ and $alpha_2$ are the roots of the characteristic polynomial $4 alpha^2 + 5 alpha + 3 = 0$, and $A_pm$ and $B_pm$ are as-yet undetermined coefficients.

We now integrate the original ODE over the region $x = [-epsilon, epsilon]$:
$$
4 left[ y' right]_{-epsilon}^epsilon + 5 left[ y right]_{-epsilon}^epsilon + int_{-epsilon}^{epsilon} (3y + 2) , dx = 8 left[ h' right]_{-epsilon}^epsilon + 7 left[ h right]_{-epsilon}^epsilon + 6epsilon
$$
In the limit as $epsilon to 0$, this becomes
$$
4 left[ y' right]_{0_-}^{0_+} + 5 left[ y right]_{0_-}^{0_+} = 7.
$$
This leads to a condition on the unknown coefficients $A_pm$ and $B_pm$. Similarly, integrating the equation twice over the same region and taking the $epsilon to 0$ limit yields
$$
4 left[ y right]_{0_-}^{0_+} = 8.
$$

The two discontinuity equations above give us two equations for the three unknowns $A_pm$ and $B_pm$. The other two equations are determined by imposing appropriate conditions as $x to -infty$; for example, if it must approach a constant, then we must have $A_- = B_- = 0$.

share|cite|improve this answer

answered Mar 13 at 19:09

Michael SeifertMichael Seifert

5,062625

$endgroup$

add a comment | 

$begingroup$

We can solve an equation like this by solving the equation in "well-behaved" regions of the real line and then integrating the ODE in a region around the singular points to find matching boundary conditions. I'll sketch out how this is done here and leave the details as an exercise for the reader.

Consider the equation
$$
4y''+5y'+3y+2=8h''+7h'+6h
$$
The right-hand side of this equation is equal to $6$ when $x > 0$, and equal to $0$ when $x < 0$. In these regions, we can therefore write
$$
4 y'' + 5 y' + 3y = begin{cases} 4 & x > 0 \ -2 & x < 0 end{cases}
$$
The solutions to this equation will then be
$$
y(t) = begin{cases} A_+ e^{alpha_1 x} + B_+ e^{alpha_2 x} + frac{4}{3} & x> 0 \ A_- e^{alpha_1 x} + B_- e^{alpha_2 x} - frac{2}{3} & x> 0
end{cases},
$$
where $alpha_1$ and $alpha_2$ are the roots of the characteristic polynomial $4 alpha^2 + 5 alpha + 3 = 0$, and $A_pm$ and $B_pm$ are as-yet undetermined coefficients.

We now integrate the original ODE over the region $x = [-epsilon, epsilon]$:
$$
4 left[ y' right]_{-epsilon}^epsilon + 5 left[ y right]_{-epsilon}^epsilon + int_{-epsilon}^{epsilon} (3y + 2) , dx = 8 left[ h' right]_{-epsilon}^epsilon + 7 left[ h right]_{-epsilon}^epsilon + 6epsilon
$$
In the limit as $epsilon to 0$, this becomes
$$
4 left[ y' right]_{0_-}^{0_+} + 5 left[ y right]_{0_-}^{0_+} = 7.
$$
This leads to a condition on the unknown coefficients $A_pm$ and $B_pm$. Similarly, integrating the equation twice over the same region and taking the $epsilon to 0$ limit yields
$$
4 left[ y right]_{0_-}^{0_+} = 8.
$$

The two discontinuity equations above give us two equations for the three unknowns $A_pm$ and $B_pm$. The other two equations are determined by imposing appropriate conditions as $x to -infty$; for example, if it must approach a constant, then we must have $A_- = B_- = 0$.

share|cite|improve this answer

answered Mar 13 at 19:09

Michael SeifertMichael Seifert

5,062625

$endgroup$

share|cite|improve this answer

answered Mar 13 at 19:09

Michael SeifertMichael Seifert

5,062625

answered Mar 13 at 19:09

Michael SeifertMichael Seifert

5,062625

answered Mar 13 at 19:09

Michael SeifertMichael Seifert

5,062625