Solve Differential Equation 2: $fracd^2ydx^2-2xfracdydx+2y=0$ [closed]

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solve $fracd^2ydx^2-2xfracdydx+2y=0$
I tried to solve through linear differential equation but it has variable with x in first derivative .

differential-equations

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edited Jul 20 at 18:57

Nosrati

19.5k41544

asked Jul 20 at 18:55

ravi yadav

295

closed as off-topic by Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus Jul 21 at 0:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus

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up vote
-2
down vote

favorite

1

solve $fracd^2ydx^2-2xfracdydx+2y=0$
I tried to solve through linear differential equation but it has variable with x in first derivative .

differential-equations

share|cite|improve this question

edited Jul 20 at 18:57

Nosrati

19.5k41544

asked Jul 20 at 18:55

ravi yadav

295

closed as off-topic by Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus Jul 21 at 0:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
-2
down vote

favorite

1

up vote
-2
down vote

favorite

1

1

solve $fracd^2ydx^2-2xfracdydx+2y=0$
I tried to solve through linear differential equation but it has variable with x in first derivative .

differential-equations

share|cite|improve this question

edited Jul 20 at 18:57

Nosrati

19.5k41544

asked Jul 20 at 18:55

ravi yadav

295

solve $fracd^2ydx^2-2xfracdydx+2y=0$
I tried to solve through linear differential equation but it has variable with x in first derivative .

differential-equations

share|cite|improve this question

edited Jul 20 at 18:57

Nosrati

19.5k41544

asked Jul 20 at 18:55

ravi yadav

295

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 18:57

Nosrati

19.5k41544

edited Jul 20 at 18:57

Nosrati

19.5k41544

edited Jul 20 at 18:57

Nosrati

19.5k41544

19.5k41544

asked Jul 20 at 18:55

ravi yadav

295

asked Jul 20 at 18:55

ravi yadav

295

asked Jul 20 at 18:55

ravi yadav

295

295

closed as off-topic by Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus Jul 21 at 0:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus Jul 21 at 0:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Shaun, Adrian Keister, Taroccoesbrocco, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

Substituting $$y=xv$$ then we get $$y'=xv'+v$$ and $$y''=xv''+2v'$$ then our equation will be

$$xv''-2v'(x^2-1)=0$$ now we Substitute

$$v'=u$$ then we get $$fracu'u=-frac2x+2x
$$

and we get

$$intfracu'udx=intleft(-frac2x+2xright)dx$$

this gives

$$log(u(x))=x^2-log(x)+C$$
Can you finish?

share|cite|improve this answer

answered Jul 20 at 19:10

Dr. Sonnhard Graubner

66.8k32659

• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes , thank you very much
  – ravi yadav
  Jul 20 at 19:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Then good luck Mr. Ravi (like Ravi Substitution?)
  – Dr. Sonnhard Graubner
  Jul 20 at 19:13
  

  
  
  
  

add a comment | 

up vote
2
down vote

$y=c_1x$ is a solution.

You can try to integrate the differential by reduction of order

Try

$$y_1=v(x)x$$
The original equation
$$y''-2xy'+2y=0$$
becomes
$$v''x+2v'(1-x^2)=0$$
Substitute $w=v'$
$$w'x+2w(1-x^2)=0$$

It's a first order,differential equation
$$(ln w)'=2x-frac 2x$$
$$ln w=x^2-2ln x + K$$
$$w=C_1frac e^x^2x^2$$
$$v'=C_1frac e^x^2x^2$$
$$v(x)=C_1int frac e^x^2x^2dx+C_2$$
$$y(x)=C_1xint frac e^x^2x^2dx+C_2x$$
$$boxedy(x)=C_1(xsqrt pitext erfi(x)-e^x^2)+C_2x$$

share|cite|improve this answer

edited Jul 20 at 19:52

answered Jul 20 at 19:07

Isham

10.6k3829

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Substituting $$y=xv$$ then we get $$y'=xv'+v$$ and $$y''=xv''+2v'$$ then our equation will be

$$xv''-2v'(x^2-1)=0$$ now we Substitute

$$v'=u$$ then we get $$fracu'u=-frac2x+2x
$$

and we get

$$intfracu'udx=intleft(-frac2x+2xright)dx$$

this gives

$$log(u(x))=x^2-log(x)+C$$
Can you finish?

share|cite|improve this answer

answered Jul 20 at 19:10

Dr. Sonnhard Graubner

66.8k32659

• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes , thank you very much
  – ravi yadav
  Jul 20 at 19:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Then good luck Mr. Ravi (like Ravi Substitution?)
  – Dr. Sonnhard Graubner
  Jul 20 at 19:13
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Substituting $$y=xv$$ then we get $$y'=xv'+v$$ and $$y''=xv''+2v'$$ then our equation will be

$$xv''-2v'(x^2-1)=0$$ now we Substitute

$$v'=u$$ then we get $$fracu'u=-frac2x+2x
$$

and we get

$$intfracu'udx=intleft(-frac2x+2xright)dx$$

this gives

$$log(u(x))=x^2-log(x)+C$$
Can you finish?

share|cite|improve this answer

answered Jul 20 at 19:10

Dr. Sonnhard Graubner

66.8k32659

• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes , thank you very much
  – ravi yadav
  Jul 20 at 19:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Then good luck Mr. Ravi (like Ravi Substitution?)
  – Dr. Sonnhard Graubner
  Jul 20 at 19:13
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Substituting $$y=xv$$ then we get $$y'=xv'+v$$ and $$y''=xv''+2v'$$ then our equation will be

$$xv''-2v'(x^2-1)=0$$ now we Substitute

$$v'=u$$ then we get $$fracu'u=-frac2x+2x
$$

and we get

$$intfracu'udx=intleft(-frac2x+2xright)dx$$

this gives

$$log(u(x))=x^2-log(x)+C$$
Can you finish?

share|cite|improve this answer

answered Jul 20 at 19:10

Dr. Sonnhard Graubner

66.8k32659

Substituting $$y=xv$$ then we get $$y'=xv'+v$$ and $$y''=xv''+2v'$$ then our equation will be

$$xv''-2v'(x^2-1)=0$$ now we Substitute

$$v'=u$$ then we get $$fracu'u=-frac2x+2x
$$

and we get

$$intfracu'udx=intleft(-frac2x+2xright)dx$$

this gives

$$log(u(x))=x^2-log(x)+C$$
Can you finish?

share|cite|improve this answer

answered Jul 20 at 19:10

Dr. Sonnhard Graubner

66.8k32659

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 19:10

Dr. Sonnhard Graubner

66.8k32659

answered Jul 20 at 19:10

Dr. Sonnhard Graubner

66.8k32659

answered Jul 20 at 19:10

Dr. Sonnhard Graubner

66.8k32659

66.8k32659

• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes , thank you very much
  – ravi yadav
  Jul 20 at 19:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Then good luck Mr. Ravi (like Ravi Substitution?)
  – Dr. Sonnhard Graubner
  Jul 20 at 19:13
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes , thank you very much
  – ravi yadav
  Jul 20 at 19:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Then good luck Mr. Ravi (like Ravi Substitution?)
  – Dr. Sonnhard Graubner
  Jul 20 at 19:13
  

  
  
  
  

yes , thank you very much
– ravi yadav
Jul 20 at 19:12

yes , thank you very much
– ravi yadav
Jul 20 at 19:12

1

1

Then good luck Mr. Ravi (like Ravi Substitution?)
– Dr. Sonnhard Graubner
Jul 20 at 19:13

Then good luck Mr. Ravi (like Ravi Substitution?)
– Dr. Sonnhard Graubner
Jul 20 at 19:13

add a comment | 

up vote
2
down vote

$y=c_1x$ is a solution.

You can try to integrate the differential by reduction of order

Try

$$y_1=v(x)x$$
The original equation
$$y''-2xy'+2y=0$$
becomes
$$v''x+2v'(1-x^2)=0$$
Substitute $w=v'$
$$w'x+2w(1-x^2)=0$$

It's a first order,differential equation
$$(ln w)'=2x-frac 2x$$
$$ln w=x^2-2ln x + K$$
$$w=C_1frac e^x^2x^2$$
$$v'=C_1frac e^x^2x^2$$
$$v(x)=C_1int frac e^x^2x^2dx+C_2$$
$$y(x)=C_1xint frac e^x^2x^2dx+C_2x$$
$$boxedy(x)=C_1(xsqrt pitext erfi(x)-e^x^2)+C_2x$$

share|cite|improve this answer

edited Jul 20 at 19:52

answered Jul 20 at 19:07

Isham

10.6k3829

add a comment | 

up vote
2
down vote

$y=c_1x$ is a solution.

You can try to integrate the differential by reduction of order

Try

$$y_1=v(x)x$$
The original equation
$$y''-2xy'+2y=0$$
becomes
$$v''x+2v'(1-x^2)=0$$
Substitute $w=v'$
$$w'x+2w(1-x^2)=0$$

It's a first order,differential equation
$$(ln w)'=2x-frac 2x$$
$$ln w=x^2-2ln x + K$$
$$w=C_1frac e^x^2x^2$$
$$v'=C_1frac e^x^2x^2$$
$$v(x)=C_1int frac e^x^2x^2dx+C_2$$
$$y(x)=C_1xint frac e^x^2x^2dx+C_2x$$
$$boxedy(x)=C_1(xsqrt pitext erfi(x)-e^x^2)+C_2x$$

share|cite|improve this answer

edited Jul 20 at 19:52

answered Jul 20 at 19:07

Isham

10.6k3829

add a comment | 

up vote
2
down vote

up vote
2
down vote

$y=c_1x$ is a solution.

You can try to integrate the differential by reduction of order

Try

$$y_1=v(x)x$$
The original equation
$$y''-2xy'+2y=0$$
becomes
$$v''x+2v'(1-x^2)=0$$
Substitute $w=v'$
$$w'x+2w(1-x^2)=0$$

It's a first order,differential equation
$$(ln w)'=2x-frac 2x$$
$$ln w=x^2-2ln x + K$$
$$w=C_1frac e^x^2x^2$$
$$v'=C_1frac e^x^2x^2$$
$$v(x)=C_1int frac e^x^2x^2dx+C_2$$
$$y(x)=C_1xint frac e^x^2x^2dx+C_2x$$
$$boxedy(x)=C_1(xsqrt pitext erfi(x)-e^x^2)+C_2x$$

share|cite|improve this answer

edited Jul 20 at 19:52

answered Jul 20 at 19:07

Isham

10.6k3829

$y=c_1x$ is a solution.

You can try to integrate the differential by reduction of order

Try

$$y_1=v(x)x$$
The original equation
$$y''-2xy'+2y=0$$
becomes
$$v''x+2v'(1-x^2)=0$$
Substitute $w=v'$
$$w'x+2w(1-x^2)=0$$

It's a first order,differential equation
$$(ln w)'=2x-frac 2x$$
$$ln w=x^2-2ln x + K$$
$$w=C_1frac e^x^2x^2$$
$$v'=C_1frac e^x^2x^2$$
$$v(x)=C_1int frac e^x^2x^2dx+C_2$$
$$y(x)=C_1xint frac e^x^2x^2dx+C_2x$$
$$boxedy(x)=C_1(xsqrt pitext erfi(x)-e^x^2)+C_2x$$

share|cite|improve this answer

edited Jul 20 at 19:52

answered Jul 20 at 19:07

Isham

10.6k3829

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 19:52

edited Jul 20 at 19:52

edited Jul 20 at 19:52

answered Jul 20 at 19:07

Isham

10.6k3829

answered Jul 20 at 19:07

Isham

10.6k3829

answered Jul 20 at 19:07

Isham

10.6k3829

10.6k3829

add a comment | 

add a comment |