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How to find the general solution to this linear ODE?

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0

$begingroup$

$$y''-frac{2y}{x^2} = 2x$$

I do not know how to solve this equation as there is $x^2$ on the left hand side. should I take $frac{2}{x^2}$ as one of variable?

ordinary-differential-equations

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edited Mar 10 at 9:58

Rodrigo de Azevedo

13k41960

asked Mar 10 at 2:11

nurul hidayahnurul hidayah

1

New contributor

nurul hidayah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know how to use the integrating factor?
  $endgroup$
  – Dylan
  Mar 10 at 3:18
  

  
  
  
  

add a comment | 

0

$begingroup$

$$y''-frac{2y}{x^2} = 2x$$

I do not know how to solve this equation as there is $x^2$ on the left hand side. should I take $frac{2}{x^2}$ as one of variable?

ordinary-differential-equations

share|cite|improve this question

edited Mar 10 at 9:58

Rodrigo de Azevedo

13k41960

asked Mar 10 at 2:11

nurul hidayahnurul hidayah

1

New contributor

nurul hidayah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know how to use the integrating factor?
  $endgroup$
  – Dylan
  Mar 10 at 3:18
  

  
  
  
  

add a comment | 

$begingroup$

$$y''-frac{2y}{x^2} = 2x$$

I do not know how to solve this equation as there is $x^2$ on the left hand side. should I take $frac{2}{x^2}$ as one of variable?

ordinary-differential-equations

share|cite|improve this question

edited Mar 10 at 9:58

Rodrigo de Azevedo

13k41960

asked Mar 10 at 2:11

nurul hidayahnurul hidayah

1

New contributor

nurul hidayah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 10 at 9:58

Rodrigo de Azevedo

13k41960

asked Mar 10 at 2:11

nurul hidayahnurul hidayah

1

New contributor

nurul hidayah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 10 at 9:58

Rodrigo de Azevedo

13k41960

asked Mar 10 at 2:11

nurul hidayahnurul hidayah

1

New contributor

nurul hidayah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 10 at 9:58

Rodrigo de Azevedo

13k41960

edited Mar 10 at 9:58

Rodrigo de Azevedo

13k41960

asked Mar 10 at 2:11

nurul hidayahnurul hidayah

1

New contributor

nurul hidayah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 10 at 2:11

nurul hidayahnurul hidayah

1

2 Answers
2

active

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2

$begingroup$

HINT

To begin with, notice that

begin{align*}
x^{2}y^{primeprime} - 2y = 2x^{3} Longleftrightarrow (x^{2}y^{primeprime} + 2xy^{prime}) - (2xy^{prime} + 2y) = 2x^{3} Longleftrightarrow (x^{2}y^{prime})^{prime} - (2xy)^{prime} = 2x^{3}
end{align*}

Then integrate both sides. Can you proceed from here?

share|cite|improve this answer

answered Mar 10 at 2:39

APC89APC89

2,341520

$endgroup$

add a comment | 

0

$begingroup$

After multiplying with $x^2$, the homogeneous equation is of Euler-Cauchy type. The basis solutions of the form $x^r$ have to satisfy $0=r(r-1)-2=(r-2)(r+1)$. Then use the method of undetermined coefficients or variation of constants to find a particular solution.

share|cite|improve this answer

answered Mar 10 at 9:55

LutzLLutzL

59.6k42057

$endgroup$

add a comment | 

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2

$begingroup$

HINT

To begin with, notice that

begin{align*}
x^{2}y^{primeprime} - 2y = 2x^{3} Longleftrightarrow (x^{2}y^{primeprime} + 2xy^{prime}) - (2xy^{prime} + 2y) = 2x^{3} Longleftrightarrow (x^{2}y^{prime})^{prime} - (2xy)^{prime} = 2x^{3}
end{align*}

Then integrate both sides. Can you proceed from here?

share|cite|improve this answer

answered Mar 10 at 2:39

APC89APC89

2,341520

$endgroup$

add a comment | 

2

$begingroup$

HINT

To begin with, notice that

begin{align*}
x^{2}y^{primeprime} - 2y = 2x^{3} Longleftrightarrow (x^{2}y^{primeprime} + 2xy^{prime}) - (2xy^{prime} + 2y) = 2x^{3} Longleftrightarrow (x^{2}y^{prime})^{prime} - (2xy)^{prime} = 2x^{3}
end{align*}

Then integrate both sides. Can you proceed from here?

share|cite|improve this answer

answered Mar 10 at 2:39

APC89APC89

2,341520

$endgroup$

add a comment | 

$begingroup$

HINT

To begin with, notice that

begin{align*}
x^{2}y^{primeprime} - 2y = 2x^{3} Longleftrightarrow (x^{2}y^{primeprime} + 2xy^{prime}) - (2xy^{prime} + 2y) = 2x^{3} Longleftrightarrow (x^{2}y^{prime})^{prime} - (2xy)^{prime} = 2x^{3}
end{align*}

Then integrate both sides. Can you proceed from here?

share|cite|improve this answer

answered Mar 10 at 2:39

APC89APC89

2,341520

$endgroup$

share|cite|improve this answer

answered Mar 10 at 2:39

APC89APC89

2,341520

answered Mar 10 at 2:39

APC89APC89

2,341520

answered Mar 10 at 2:39

APC89APC89

2,341520

0

$begingroup$

After multiplying with $x^2$, the homogeneous equation is of Euler-Cauchy type. The basis solutions of the form $x^r$ have to satisfy $0=r(r-1)-2=(r-2)(r+1)$. Then use the method of undetermined coefficients or variation of constants to find a particular solution.

share|cite|improve this answer

answered Mar 10 at 9:55

LutzLLutzL

59.6k42057

$endgroup$

add a comment | 

0

$begingroup$

After multiplying with $x^2$, the homogeneous equation is of Euler-Cauchy type. The basis solutions of the form $x^r$ have to satisfy $0=r(r-1)-2=(r-2)(r+1)$. Then use the method of undetermined coefficients or variation of constants to find a particular solution.

share|cite|improve this answer

answered Mar 10 at 9:55

LutzLLutzL

59.6k42057

$endgroup$

add a comment | 

$begingroup$

After multiplying with $x^2$, the homogeneous equation is of Euler-Cauchy type. The basis solutions of the form $x^r$ have to satisfy $0=r(r-1)-2=(r-2)(r+1)$. Then use the method of undetermined coefficients or variation of constants to find a particular solution.

share|cite|improve this answer

answered Mar 10 at 9:55

LutzLLutzL

59.6k42057

$endgroup$

share|cite|improve this answer

answered Mar 10 at 9:55

LutzLLutzL

59.6k42057

answered Mar 10 at 9:55

LutzLLutzL

59.6k42057

answered Mar 10 at 9:55

LutzLLutzL

59.6k42057