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Solving Initial Value Problems [on hold]

0

$begingroup$

Considering the following IVP

$y’= frac11+x^2 -2y^2 hspace2cm x in (0,1]$

$y(0)=0$

I need to show that the solution is $ fracx1+x^2$

ordinary-differential-equations initial-value-problems

share|cite|improve this question

edited Apr 10 at 14:29

Saad

20.6k92452

asked Apr 8 at 20:08

user661721

$endgroup$

put on hold as off-topic by José Carlos Santos, Saad, Strants, T. Bongers, Lee David Chung Lin Apr 11 at 0:22

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." – Saad, Strants, T. Bongers, Lee David Chung Lin

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What IVP? Please include your problem.
  $endgroup$
  – Paras Khosla
  Apr 10 at 14:04
  

  
  
  
  

add a comment | 

0

$begingroup$

Considering the following IVP

$y’= frac11+x^2 -2y^2 hspace2cm x in (0,1]$

$y(0)=0$

I need to show that the solution is $ fracx1+x^2$

ordinary-differential-equations initial-value-problems

share|cite|improve this question

edited Apr 10 at 14:29

Saad

20.6k92452

asked Apr 8 at 20:08

user661721

$endgroup$

put on hold as off-topic by José Carlos Santos, Saad, Strants, T. Bongers, Lee David Chung Lin Apr 11 at 0:22

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." – Saad, Strants, T. Bongers, Lee David Chung Lin

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What IVP? Please include your problem.
  $endgroup$
  – Paras Khosla
  Apr 10 at 14:04
  

  
  
  
  

add a comment | 

$begingroup$

Considering the following IVP

$y’= frac11+x^2 -2y^2 hspace2cm x in (0,1]$

$y(0)=0$

I need to show that the solution is $ fracx1+x^2$

ordinary-differential-equations initial-value-problems

share|cite|improve this question

edited Apr 10 at 14:29

Saad

20.6k92452

asked Apr 8 at 20:08

user661721

$endgroup$

share|cite|improve this question

edited Apr 10 at 14:29

Saad

20.6k92452

asked Apr 8 at 20:08

user661721

share|cite|improve this question

edited Apr 10 at 14:29

Saad

20.6k92452

asked Apr 8 at 20:08

user661721

edited Apr 10 at 14:29

Saad

20.6k92452

edited Apr 10 at 14:29

Saad

20.6k92452

1 Answer
1

active

oldest

votes

2

$begingroup$

Hint: Checking $y(0)=0/(1+0^2)=0$ shows that the initial value is correct. Then plug

$$y(x) = dfracx1+x^2$$ into the differential equaiton and check if the equation is fullfilled.

EDIT: We know that

$$y'(x) = dfrac1-x^2(1+x^2)^2.$$
At the same time the right hand side evaluates to

$$ dfrac11+x^2-2y^2=dfrac11+x^2-2dfracx^2(1+x^2)^2=dfrac1-x^2(1+x^2)^2.$$

Hence, the left side equals the right side which verifies that $y(x) = x/(1+x^2)$ is a solution to the differential equation.

share|cite|improve this answer

edited Apr 9 at 15:31

answered Apr 8 at 20:12

MachineLearnerMachineLearner

1,481212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1. I like your straight answer.
  $endgroup$
  – Felix Marin
  Apr 8 at 21:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really appreciate your feedback :).
  $endgroup$
  – MachineLearner
  Apr 9 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MachineLearner i checked the intial condition and then i have tried plugging it into the equation and i managed to get $ y'= frac1-x^2(1+x^2)^2$ . Im unsure how to check if this equation is fulfilled
  $endgroup$
  – user661721
  Apr 9 at 12:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathwizard: See my edit.
  $endgroup$
  – MachineLearner
  Apr 9 at 15:31
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint: Checking $y(0)=0/(1+0^2)=0$ shows that the initial value is correct. Then plug

$$y(x) = dfracx1+x^2$$ into the differential equaiton and check if the equation is fullfilled.

EDIT: We know that

$$y'(x) = dfrac1-x^2(1+x^2)^2.$$
At the same time the right hand side evaluates to

$$ dfrac11+x^2-2y^2=dfrac11+x^2-2dfracx^2(1+x^2)^2=dfrac1-x^2(1+x^2)^2.$$

Hence, the left side equals the right side which verifies that $y(x) = x/(1+x^2)$ is a solution to the differential equation.

share|cite|improve this answer

edited Apr 9 at 15:31

answered Apr 8 at 20:12

MachineLearnerMachineLearner

1,481212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1. I like your straight answer.
  $endgroup$
  – Felix Marin
  Apr 8 at 21:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really appreciate your feedback :).
  $endgroup$
  – MachineLearner
  Apr 9 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MachineLearner i checked the intial condition and then i have tried plugging it into the equation and i managed to get $ y'= frac1-x^2(1+x^2)^2$ . Im unsure how to check if this equation is fulfilled
  $endgroup$
  – user661721
  Apr 9 at 12:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathwizard: See my edit.
  $endgroup$
  – MachineLearner
  Apr 9 at 15:31
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint: Checking $y(0)=0/(1+0^2)=0$ shows that the initial value is correct. Then plug

$$y(x) = dfracx1+x^2$$ into the differential equaiton and check if the equation is fullfilled.

EDIT: We know that

$$y'(x) = dfrac1-x^2(1+x^2)^2.$$
At the same time the right hand side evaluates to

$$ dfrac11+x^2-2y^2=dfrac11+x^2-2dfracx^2(1+x^2)^2=dfrac1-x^2(1+x^2)^2.$$

Hence, the left side equals the right side which verifies that $y(x) = x/(1+x^2)$ is a solution to the differential equation.

share|cite|improve this answer

edited Apr 9 at 15:31

answered Apr 8 at 20:12

MachineLearnerMachineLearner

1,481212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1. I like your straight answer.
  $endgroup$
  – Felix Marin
  Apr 8 at 21:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really appreciate your feedback :).
  $endgroup$
  – MachineLearner
  Apr 9 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MachineLearner i checked the intial condition and then i have tried plugging it into the equation and i managed to get $ y'= frac1-x^2(1+x^2)^2$ . Im unsure how to check if this equation is fulfilled
  $endgroup$
  – user661721
  Apr 9 at 12:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathwizard: See my edit.
  $endgroup$
  – MachineLearner
  Apr 9 at 15:31
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Checking $y(0)=0/(1+0^2)=0$ shows that the initial value is correct. Then plug

$$y(x) = dfracx1+x^2$$ into the differential equaiton and check if the equation is fullfilled.

EDIT: We know that

$$y'(x) = dfrac1-x^2(1+x^2)^2.$$
At the same time the right hand side evaluates to

$$ dfrac11+x^2-2y^2=dfrac11+x^2-2dfracx^2(1+x^2)^2=dfrac1-x^2(1+x^2)^2.$$

Hence, the left side equals the right side which verifies that $y(x) = x/(1+x^2)$ is a solution to the differential equation.

share|cite|improve this answer

edited Apr 9 at 15:31

answered Apr 8 at 20:12

MachineLearnerMachineLearner

1,481212

$endgroup$

share|cite|improve this answer

edited Apr 9 at 15:31

answered Apr 8 at 20:12

MachineLearnerMachineLearner

1,481212

answered Apr 8 at 20:12

MachineLearnerMachineLearner

1,481212

answered Apr 8 at 20:12

MachineLearnerMachineLearner

1,481212