Second order differential equation with variation of parameters

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I'd like to solve $y''+frac2xy'=frac1x^2$

I identify this as Euler-Cauchy and rewrite to $x^2y''+2xy'=1$

So I have a homogenous solution of $y_h=C_1+fracC_2x$

Second order Variation of parameters:

$C_1(x)'=frac1frac1xfrac1x^2=x rightarrow C_1 = fracx^22$

$C_1(x)'=-frac1cdot1frac1x^2=-x^2 rightarrow C_2 = -fracx^33$

Is this procedure correct?

calculus differential-equations

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asked Jul 20 at 8:52

user83855

31

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up vote
0
down vote

favorite

I'd like to solve $y''+frac2xy'=frac1x^2$

I identify this as Euler-Cauchy and rewrite to $x^2y''+2xy'=1$

So I have a homogenous solution of $y_h=C_1+fracC_2x$

Second order Variation of parameters:

$C_1(x)'=frac1frac1xfrac1x^2=x rightarrow C_1 = fracx^22$

$C_1(x)'=-frac1cdot1frac1x^2=-x^2 rightarrow C_2 = -fracx^33$

Is this procedure correct?

calculus differential-equations

share|cite|improve this question

asked Jul 20 at 8:52

user83855

31

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I'd like to solve $y''+frac2xy'=frac1x^2$

I identify this as Euler-Cauchy and rewrite to $x^2y''+2xy'=1$

So I have a homogenous solution of $y_h=C_1+fracC_2x$

Second order Variation of parameters:

$C_1(x)'=frac1frac1xfrac1x^2=x rightarrow C_1 = fracx^22$

$C_1(x)'=-frac1cdot1frac1x^2=-x^2 rightarrow C_2 = -fracx^33$

Is this procedure correct?

calculus differential-equations

share|cite|improve this question

asked Jul 20 at 8:52

user83855

31

I'd like to solve $y''+frac2xy'=frac1x^2$

I identify this as Euler-Cauchy and rewrite to $x^2y''+2xy'=1$

So I have a homogenous solution of $y_h=C_1+fracC_2x$

Second order Variation of parameters:

$C_1(x)'=frac1frac1xfrac1x^2=x rightarrow C_1 = fracx^22$

$C_1(x)'=-frac1cdot1frac1x^2=-x^2 rightarrow C_2 = -fracx^33$

Is this procedure correct?

calculus differential-equations

share|cite|improve this question

asked Jul 20 at 8:52

user83855

31

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 8:52

user83855

31

asked Jul 20 at 8:52

user83855

31

asked Jul 20 at 8:52

user83855

31

31

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
0
down vote



accepted

No. You can see that $C_1+frac C_2 x$ is not a solution to the DE for the values of $C_1,C_2$ you have computed.. What you have to do is compute first and second derivatives of $C_1+frac C_2 x$, plug them in to the DE. There will be some cancellations and you will get a first order DE in $C_1'$ and $C_2'$. Solving these and integrating you can find the solutions.

share|cite|improve this answer

answered Jul 20 at 9:05

Kavi Rama Murthy

20.6k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can I not just use en.wikipedia.org/wiki/… (which is what you are proposing right?) the "result", where $A'(x)=-frac1Wu_2(x)f(x)$ an integrate this? Or is the problem here, that my EQ is not of the form $y''+p(x)y'+q(x)y=f(x)$?
  – user83855
  Jul 20 at 9:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use the Wikipedia result. Basically I have suggested the same approach in my answer.
  – Kavi Rama Murthy
  Jul 20 at 9:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  just to make sure: in my case $u_1(x)=1$ and $u_2(x)=1/x$, right?
  – user83855
  Jul 20 at 11:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, those are $u_1$ and $u_2$.
  – Kavi Rama Murthy
  Jul 20 at 11:54
  

  
  
  
  

add a comment | 

up vote
1
down vote

By using $x=e^z$ and $D_1=d/dz$, the equation becomes. $$(D_1(D_1-1)+2D_1)y=1$$
So the particular integral is $$frac1D_1(D_1+1)1=frac1D_11=z=log x$$
So the solution is $$y=C_1+fracC_2x+log x$$

share|cite|improve this answer

answered Jul 20 at 9:02

Piyush Divyanakar

3,258122

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am sorry not familiar with method of variation of parameters.
  – Piyush Divyanakar
  Jul 20 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The title says OP wants to solve the equation by variation of parameters.
  – Kavi Rama Murthy
  Jul 20 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 nice method.......@PiyushDivyanakar
  – Isham
  Jul 20 at 13:47
  

  
  
  
  

add a comment | 

up vote
0
down vote

$$y''+frac2xy'=frac1x^2$$
Substitute $z=y'$
$$x^2z'+2xz=1$$
Using variation of parameter

Solving the homogeneous equation
$$x^2z'+2xz=0$$
$$(x^2z)'=0 implies x^2z=K_1$$
$$z=frac K_1x^2$$
$$x^2z'+2xz=1$$
$$x^2(-2K_1x^-3)+K'_1+2xfrac K_1x^2=1$$
$$K'_1=1$$
$$K_1=x+C$$
$$implies z=frac x+Cx^2$$
$$y'=frac x+Cx^2$$
Simply integrate
$$y=int frac x+Cx^2dx$$
$$y=ln|x|+frac C_1x+C_2$$

share|cite|improve this answer

answered Jul 20 at 9:36

Isham

10.6k3829

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

No. You can see that $C_1+frac C_2 x$ is not a solution to the DE for the values of $C_1,C_2$ you have computed.. What you have to do is compute first and second derivatives of $C_1+frac C_2 x$, plug them in to the DE. There will be some cancellations and you will get a first order DE in $C_1'$ and $C_2'$. Solving these and integrating you can find the solutions.

share|cite|improve this answer

answered Jul 20 at 9:05

Kavi Rama Murthy

20.6k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can I not just use en.wikipedia.org/wiki/… (which is what you are proposing right?) the "result", where $A'(x)=-frac1Wu_2(x)f(x)$ an integrate this? Or is the problem here, that my EQ is not of the form $y''+p(x)y'+q(x)y=f(x)$?
  – user83855
  Jul 20 at 9:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use the Wikipedia result. Basically I have suggested the same approach in my answer.
  – Kavi Rama Murthy
  Jul 20 at 9:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  just to make sure: in my case $u_1(x)=1$ and $u_2(x)=1/x$, right?
  – user83855
  Jul 20 at 11:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, those are $u_1$ and $u_2$.
  – Kavi Rama Murthy
  Jul 20 at 11:54
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

No. You can see that $C_1+frac C_2 x$ is not a solution to the DE for the values of $C_1,C_2$ you have computed.. What you have to do is compute first and second derivatives of $C_1+frac C_2 x$, plug them in to the DE. There will be some cancellations and you will get a first order DE in $C_1'$ and $C_2'$. Solving these and integrating you can find the solutions.

share|cite|improve this answer

answered Jul 20 at 9:05

Kavi Rama Murthy

20.6k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can I not just use en.wikipedia.org/wiki/… (which is what you are proposing right?) the "result", where $A'(x)=-frac1Wu_2(x)f(x)$ an integrate this? Or is the problem here, that my EQ is not of the form $y''+p(x)y'+q(x)y=f(x)$?
  – user83855
  Jul 20 at 9:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use the Wikipedia result. Basically I have suggested the same approach in my answer.
  – Kavi Rama Murthy
  Jul 20 at 9:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  just to make sure: in my case $u_1(x)=1$ and $u_2(x)=1/x$, right?
  – user83855
  Jul 20 at 11:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, those are $u_1$ and $u_2$.
  – Kavi Rama Murthy
  Jul 20 at 11:54
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

No. You can see that $C_1+frac C_2 x$ is not a solution to the DE for the values of $C_1,C_2$ you have computed.. What you have to do is compute first and second derivatives of $C_1+frac C_2 x$, plug them in to the DE. There will be some cancellations and you will get a first order DE in $C_1'$ and $C_2'$. Solving these and integrating you can find the solutions.

share|cite|improve this answer

answered Jul 20 at 9:05

Kavi Rama Murthy

20.6k2830

No. You can see that $C_1+frac C_2 x$ is not a solution to the DE for the values of $C_1,C_2$ you have computed.. What you have to do is compute first and second derivatives of $C_1+frac C_2 x$, plug them in to the DE. There will be some cancellations and you will get a first order DE in $C_1'$ and $C_2'$. Solving these and integrating you can find the solutions.

share|cite|improve this answer

answered Jul 20 at 9:05

Kavi Rama Murthy

20.6k2830

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 9:05

Kavi Rama Murthy

20.6k2830

answered Jul 20 at 9:05

Kavi Rama Murthy

20.6k2830

answered Jul 20 at 9:05

Kavi Rama Murthy

20.6k2830

20.6k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can I not just use en.wikipedia.org/wiki/… (which is what you are proposing right?) the "result", where $A'(x)=-frac1Wu_2(x)f(x)$ an integrate this? Or is the problem here, that my EQ is not of the form $y''+p(x)y'+q(x)y=f(x)$?
  – user83855
  Jul 20 at 9:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use the Wikipedia result. Basically I have suggested the same approach in my answer.
  – Kavi Rama Murthy
  Jul 20 at 9:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  just to make sure: in my case $u_1(x)=1$ and $u_2(x)=1/x$, right?
  – user83855
  Jul 20 at 11:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, those are $u_1$ and $u_2$.
  – Kavi Rama Murthy
  Jul 20 at 11:54
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can I not just use en.wikipedia.org/wiki/… (which is what you are proposing right?) the "result", where $A'(x)=-frac1Wu_2(x)f(x)$ an integrate this? Or is the problem here, that my EQ is not of the form $y''+p(x)y'+q(x)y=f(x)$?
  – user83855
  Jul 20 at 9:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use the Wikipedia result. Basically I have suggested the same approach in my answer.
  – Kavi Rama Murthy
  Jul 20 at 9:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  just to make sure: in my case $u_1(x)=1$ and $u_2(x)=1/x$, right?
  – user83855
  Jul 20 at 11:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, those are $u_1$ and $u_2$.
  – Kavi Rama Murthy
  Jul 20 at 11:54
  

  
  
  
  

Can I not just use en.wikipedia.org/wiki/… (which is what you are proposing right?) the "result", where $A'(x)=-frac1Wu_2(x)f(x)$ an integrate this? Or is the problem here, that my EQ is not of the form $y''+p(x)y'+q(x)y=f(x)$?
– user83855
Jul 20 at 9:27

Can I not just use en.wikipedia.org/wiki/… (which is what you are proposing right?) the "result", where $A'(x)=-frac1Wu_2(x)f(x)$ an integrate this? Or is the problem here, that my EQ is not of the form $y''+p(x)y'+q(x)y=f(x)$?
– user83855
Jul 20 at 9:27

You can use the Wikipedia result. Basically I have suggested the same approach in my answer.
– Kavi Rama Murthy
Jul 20 at 9:32

You can use the Wikipedia result. Basically I have suggested the same approach in my answer.
– Kavi Rama Murthy
Jul 20 at 9:32

just to make sure: in my case $u_1(x)=1$ and $u_2(x)=1/x$, right?
– user83855
Jul 20 at 11:29

just to make sure: in my case $u_1(x)=1$ and $u_2(x)=1/x$, right?
– user83855
Jul 20 at 11:29

Yes, those are $u_1$ and $u_2$.
– Kavi Rama Murthy
Jul 20 at 11:54

Yes, those are $u_1$ and $u_2$.
– Kavi Rama Murthy
Jul 20 at 11:54

add a comment | 

up vote
1
down vote

By using $x=e^z$ and $D_1=d/dz$, the equation becomes. $$(D_1(D_1-1)+2D_1)y=1$$
So the particular integral is $$frac1D_1(D_1+1)1=frac1D_11=z=log x$$
So the solution is $$y=C_1+fracC_2x+log x$$

share|cite|improve this answer

answered Jul 20 at 9:02

Piyush Divyanakar

3,258122

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am sorry not familiar with method of variation of parameters.
  – Piyush Divyanakar
  Jul 20 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The title says OP wants to solve the equation by variation of parameters.
  – Kavi Rama Murthy
  Jul 20 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 nice method.......@PiyushDivyanakar
  – Isham
  Jul 20 at 13:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

By using $x=e^z$ and $D_1=d/dz$, the equation becomes. $$(D_1(D_1-1)+2D_1)y=1$$
So the particular integral is $$frac1D_1(D_1+1)1=frac1D_11=z=log x$$
So the solution is $$y=C_1+fracC_2x+log x$$

share|cite|improve this answer

answered Jul 20 at 9:02

Piyush Divyanakar

3,258122

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am sorry not familiar with method of variation of parameters.
  – Piyush Divyanakar
  Jul 20 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The title says OP wants to solve the equation by variation of parameters.
  – Kavi Rama Murthy
  Jul 20 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 nice method.......@PiyushDivyanakar
  – Isham
  Jul 20 at 13:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

By using $x=e^z$ and $D_1=d/dz$, the equation becomes. $$(D_1(D_1-1)+2D_1)y=1$$
So the particular integral is $$frac1D_1(D_1+1)1=frac1D_11=z=log x$$
So the solution is $$y=C_1+fracC_2x+log x$$

share|cite|improve this answer

answered Jul 20 at 9:02

Piyush Divyanakar

3,258122

By using $x=e^z$ and $D_1=d/dz$, the equation becomes. $$(D_1(D_1-1)+2D_1)y=1$$
So the particular integral is $$frac1D_1(D_1+1)1=frac1D_11=z=log x$$
So the solution is $$y=C_1+fracC_2x+log x$$

share|cite|improve this answer

answered Jul 20 at 9:02

Piyush Divyanakar

3,258122

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 9:02

Piyush Divyanakar

3,258122

answered Jul 20 at 9:02

Piyush Divyanakar

3,258122

answered Jul 20 at 9:02

Piyush Divyanakar

3,258122

3,258122

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am sorry not familiar with method of variation of parameters.
  – Piyush Divyanakar
  Jul 20 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The title says OP wants to solve the equation by variation of parameters.
  – Kavi Rama Murthy
  Jul 20 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 nice method.......@PiyushDivyanakar
  – Isham
  Jul 20 at 13:47
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I am sorry not familiar with method of variation of parameters.
  – Piyush Divyanakar
  Jul 20 at 9:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The title says OP wants to solve the equation by variation of parameters.
  – Kavi Rama Murthy
  Jul 20 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 nice method.......@PiyushDivyanakar
  – Isham
  Jul 20 at 13:47
  

  
  
  
  

1

1

I am sorry not familiar with method of variation of parameters.
– Piyush Divyanakar
Jul 20 at 9:06

I am sorry not familiar with method of variation of parameters.
– Piyush Divyanakar
Jul 20 at 9:06

The title says OP wants to solve the equation by variation of parameters.
– Kavi Rama Murthy
Jul 20 at 9:07

The title says OP wants to solve the equation by variation of parameters.
– Kavi Rama Murthy
Jul 20 at 9:07

+1 nice method.......@PiyushDivyanakar
– Isham
Jul 20 at 13:47

+1 nice method.......@PiyushDivyanakar
– Isham
Jul 20 at 13:47

add a comment | 

up vote
0
down vote

$$y''+frac2xy'=frac1x^2$$
Substitute $z=y'$
$$x^2z'+2xz=1$$
Using variation of parameter

Solving the homogeneous equation
$$x^2z'+2xz=0$$
$$(x^2z)'=0 implies x^2z=K_1$$
$$z=frac K_1x^2$$
$$x^2z'+2xz=1$$
$$x^2(-2K_1x^-3)+K'_1+2xfrac K_1x^2=1$$
$$K'_1=1$$
$$K_1=x+C$$
$$implies z=frac x+Cx^2$$
$$y'=frac x+Cx^2$$
Simply integrate
$$y=int frac x+Cx^2dx$$
$$y=ln|x|+frac C_1x+C_2$$

share|cite|improve this answer

answered Jul 20 at 9:36

Isham

10.6k3829

add a comment | 

up vote
0
down vote

$$y''+frac2xy'=frac1x^2$$
Substitute $z=y'$
$$x^2z'+2xz=1$$
Using variation of parameter

Solving the homogeneous equation
$$x^2z'+2xz=0$$
$$(x^2z)'=0 implies x^2z=K_1$$
$$z=frac K_1x^2$$
$$x^2z'+2xz=1$$
$$x^2(-2K_1x^-3)+K'_1+2xfrac K_1x^2=1$$
$$K'_1=1$$
$$K_1=x+C$$
$$implies z=frac x+Cx^2$$
$$y'=frac x+Cx^2$$
Simply integrate
$$y=int frac x+Cx^2dx$$
$$y=ln|x|+frac C_1x+C_2$$

share|cite|improve this answer

answered Jul 20 at 9:36

Isham

10.6k3829

add a comment | 

up vote
0
down vote

up vote
0
down vote

$$y''+frac2xy'=frac1x^2$$
Substitute $z=y'$
$$x^2z'+2xz=1$$
Using variation of parameter

Solving the homogeneous equation
$$x^2z'+2xz=0$$
$$(x^2z)'=0 implies x^2z=K_1$$
$$z=frac K_1x^2$$
$$x^2z'+2xz=1$$
$$x^2(-2K_1x^-3)+K'_1+2xfrac K_1x^2=1$$
$$K'_1=1$$
$$K_1=x+C$$
$$implies z=frac x+Cx^2$$
$$y'=frac x+Cx^2$$
Simply integrate
$$y=int frac x+Cx^2dx$$
$$y=ln|x|+frac C_1x+C_2$$

share|cite|improve this answer

answered Jul 20 at 9:36

Isham

10.6k3829

$$y''+frac2xy'=frac1x^2$$
Substitute $z=y'$
$$x^2z'+2xz=1$$
Using variation of parameter

Solving the homogeneous equation
$$x^2z'+2xz=0$$
$$(x^2z)'=0 implies x^2z=K_1$$
$$z=frac K_1x^2$$
$$x^2z'+2xz=1$$
$$x^2(-2K_1x^-3)+K'_1+2xfrac K_1x^2=1$$
$$K'_1=1$$
$$K_1=x+C$$
$$implies z=frac x+Cx^2$$
$$y'=frac x+Cx^2$$
Simply integrate
$$y=int frac x+Cx^2dx$$
$$y=ln|x|+frac C_1x+C_2$$

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answered Jul 20 at 9:36

Isham

10.6k3829

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share|cite|improve this answer

answered Jul 20 at 9:36

Isham

10.6k3829

answered Jul 20 at 9:36

Isham

10.6k3829

answered Jul 20 at 9:36

Isham

10.6k3829

10.6k3829

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