Differential equation $y'=2xy-x^2y'; y(-3)=1$ [closed]

-1

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Differential equation $y'=2xy-x^2y', y(-3)=1$

I've got the equation: $ln(y)=ln(1+x^2)+C$ where $C=-ln(10)$ and this is incorrectly.

Thank you for your help.

ordinary-differential-equations

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asked Jan 11 at 9:05

J.DoeJ.Doe

899

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closed as off-topic by Did, Nosrati, Abcd, amWhy, Song Jan 11 at 21:56

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Abcd, amWhy

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Exponentiate both sides.
  $endgroup$
  – Claude Leibovici
  Jan 11 at 9:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much!
  $endgroup$
  – J.Doe
  Jan 11 at 9:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $ ln(y)=ln(1+x^2)+ln, C rightarrow y=C (1+x^2) $ Plug in BC
  $endgroup$
  – Narasimham
  Jan 11 at 18:17
  
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

Differential equation $y'=2xy-x^2y', y(-3)=1$

I've got the equation: $ln(y)=ln(1+x^2)+C$ where $C=-ln(10)$ and this is incorrectly.

Thank you for your help.

ordinary-differential-equations

share|cite|improve this question

asked Jan 11 at 9:05

J.DoeJ.Doe

899

$endgroup$

closed as off-topic by Did, Nosrati, Abcd, amWhy, Song Jan 11 at 21:56

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Abcd, amWhy

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Exponentiate both sides.
  $endgroup$
  – Claude Leibovici
  Jan 11 at 9:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much!
  $endgroup$
  – J.Doe
  Jan 11 at 9:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $ ln(y)=ln(1+x^2)+ln, C rightarrow y=C (1+x^2) $ Plug in BC
  $endgroup$
  – Narasimham
  Jan 11 at 18:17
  
  
  

  
  
  
  

add a comment | 

-1

-1

-1

$begingroup$

Differential equation $y'=2xy-x^2y', y(-3)=1$

I've got the equation: $ln(y)=ln(1+x^2)+C$ where $C=-ln(10)$ and this is incorrectly.

Thank you for your help.

ordinary-differential-equations

share|cite|improve this question

asked Jan 11 at 9:05

J.DoeJ.Doe

899

$endgroup$

Differential equation $y'=2xy-x^2y', y(-3)=1$

I've got the equation: $ln(y)=ln(1+x^2)+C$ where $C=-ln(10)$ and this is incorrectly.

Thank you for your help.

ordinary-differential-equations

ordinary-differential-equations

share|cite|improve this question

asked Jan 11 at 9:05

J.DoeJ.Doe

899

share|cite|improve this question

asked Jan 11 at 9:05

J.DoeJ.Doe

899

share|cite|improve this question

share|cite|improve this question

asked Jan 11 at 9:05

J.DoeJ.Doe

899

asked Jan 11 at 9:05

J.DoeJ.Doe

899

asked Jan 11 at 9:05

J.DoeJ.Doe

899

899

closed as off-topic by Did, Nosrati, Abcd, amWhy, Song Jan 11 at 21:56

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Abcd, amWhy

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

closed as off-topic by Did, Nosrati, Abcd, amWhy, Song Jan 11 at 21:56

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Abcd, amWhy

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Exponentiate both sides.
  $endgroup$
  – Claude Leibovici
  Jan 11 at 9:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much!
  $endgroup$
  – J.Doe
  Jan 11 at 9:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $ ln(y)=ln(1+x^2)+ln, C rightarrow y=C (1+x^2) $ Plug in BC
  $endgroup$
  – Narasimham
  Jan 11 at 18:17
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Exponentiate both sides.
  $endgroup$
  – Claude Leibovici
  Jan 11 at 9:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much!
  $endgroup$
  – J.Doe
  Jan 11 at 9:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $ ln(y)=ln(1+x^2)+ln, C rightarrow y=C (1+x^2) $ Plug in BC
  $endgroup$
  – Narasimham
  Jan 11 at 18:17
  
  
  

  
  
  
  

2

2

$begingroup$
Exponentiate both sides.
$endgroup$
– Claude Leibovici
Jan 11 at 9:10

$begingroup$
Exponentiate both sides.
$endgroup$
– Claude Leibovici
Jan 11 at 9:10

$begingroup$
Thank you very much!
$endgroup$
– J.Doe
Jan 11 at 9:13

$begingroup$
Thank you very much!
$endgroup$
– J.Doe
Jan 11 at 9:13

$begingroup$
$ ln(y)=ln(1+x^2)+ln, C rightarrow y=C (1+x^2) $ Plug in BC
$endgroup$
– Narasimham
Jan 11 at 18:17

$begingroup$
$ ln(y)=ln(1+x^2)+ln, C rightarrow y=C (1+x^2) $ Plug in BC
$endgroup$
– Narasimham
Jan 11 at 18:17

add a comment | 

1 Answer
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Rewrite your equation as
$$frac{dy}{dx}=frac{2xy}{1+x^2}.$$
The general solution of
$$frac{dy}{dx}=f(x)y$$
is
$$y=Ce^{int f(x)dx}$$
where $C$ is an arbitrary constant.
Here you have
$$f(x)=frac{2x}{1+x^2}$$
so the solution is
$$y=Ce^{ln(1+x^2)}=C(1+x^2)$$
and
$$y(-3)=10C=-1.$$

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answered Jan 11 at 11:33

bofbof

51.5k558120

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add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Rewrite your equation as
$$frac{dy}{dx}=frac{2xy}{1+x^2}.$$
The general solution of
$$frac{dy}{dx}=f(x)y$$
is
$$y=Ce^{int f(x)dx}$$
where $C$ is an arbitrary constant.
Here you have
$$f(x)=frac{2x}{1+x^2}$$
so the solution is
$$y=Ce^{ln(1+x^2)}=C(1+x^2)$$
and
$$y(-3)=10C=-1.$$

share|cite|improve this answer

answered Jan 11 at 11:33

bofbof

51.5k558120

$endgroup$

add a comment | 

1

$begingroup$

Rewrite your equation as
$$frac{dy}{dx}=frac{2xy}{1+x^2}.$$
The general solution of
$$frac{dy}{dx}=f(x)y$$
is
$$y=Ce^{int f(x)dx}$$
where $C$ is an arbitrary constant.
Here you have
$$f(x)=frac{2x}{1+x^2}$$
so the solution is
$$y=Ce^{ln(1+x^2)}=C(1+x^2)$$
and
$$y(-3)=10C=-1.$$

share|cite|improve this answer

answered Jan 11 at 11:33

bofbof

51.5k558120

$endgroup$

add a comment | 

1

1

1

$begingroup$

Rewrite your equation as
$$frac{dy}{dx}=frac{2xy}{1+x^2}.$$
The general solution of
$$frac{dy}{dx}=f(x)y$$
is
$$y=Ce^{int f(x)dx}$$
where $C$ is an arbitrary constant.
Here you have
$$f(x)=frac{2x}{1+x^2}$$
so the solution is
$$y=Ce^{ln(1+x^2)}=C(1+x^2)$$
and
$$y(-3)=10C=-1.$$

share|cite|improve this answer

answered Jan 11 at 11:33

bofbof

51.5k558120

$endgroup$

Rewrite your equation as
$$frac{dy}{dx}=frac{2xy}{1+x^2}.$$
The general solution of
$$frac{dy}{dx}=f(x)y$$
is
$$y=Ce^{int f(x)dx}$$
where $C$ is an arbitrary constant.
Here you have
$$f(x)=frac{2x}{1+x^2}$$
so the solution is
$$y=Ce^{ln(1+x^2)}=C(1+x^2)$$
and
$$y(-3)=10C=-1.$$

share|cite|improve this answer

answered Jan 11 at 11:33

bofbof

51.5k558120

share|cite|improve this answer

share|cite|improve this answer

answered Jan 11 at 11:33

bofbof

51.5k558120

answered Jan 11 at 11:33

bofbof

51.5k558120

answered Jan 11 at 11:33

bofbof

51.5k558120

51.5k558120

add a comment | 

add a comment | 

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