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Solving a second order nonlinear differential equation

Second order differential equations using MATLABsolving second-order nonlinear ordinary differential equationSecond order differential equationNonlinear Second Order Differential EquationSolving differential equationSolving Second Order Nonlinear Nonhomogeneous ODE (Constant Coefficients)How to solve a nonlinear second order differential equation?Solving second order differential equation 1Second-order differential equation with initial conditionsSolving Nonlinear differential equation system

1

1

$begingroup$

Solve $$y’’ + 2y = 4 -frac1sqrty$$ with initial conditions $y(0) = 1$ and $y’(0) = 2$.

I am trying different substitutions, but none seem to work.

ordinary-differential-equations

share|cite|improve this question

edited Mar 13 at 18:51

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 18:41

breakubreaku

405

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo I am trying different substitutions but it does not seem to work
  $endgroup$
  – breaku
  Mar 13 at 18:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Any ODE of the form $y''=F(y)$ can be solved (in principle) through multiplying by $y'$ and integrating once, to get a separable first order ODE.
  $endgroup$
  – Hans Lundmark
  Mar 14 at 8:10
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Solve $$y’’ + 2y = 4 -frac1sqrty$$ with initial conditions $y(0) = 1$ and $y’(0) = 2$.

I am trying different substitutions, but none seem to work.

ordinary-differential-equations

share|cite|improve this question

edited Mar 13 at 18:51

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 18:41

breakubreaku

405

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo I am trying different substitutions but it does not seem to work
  $endgroup$
  – breaku
  Mar 13 at 18:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Any ODE of the form $y''=F(y)$ can be solved (in principle) through multiplying by $y'$ and integrating once, to get a separable first order ODE.
  $endgroup$
  – Hans Lundmark
  Mar 14 at 8:10
  

  
  
  
  

add a comment | 

$begingroup$

Solve $$y’’ + 2y = 4 -frac1sqrty$$ with initial conditions $y(0) = 1$ and $y’(0) = 2$.

I am trying different substitutions, but none seem to work.

ordinary-differential-equations

share|cite|improve this question

edited Mar 13 at 18:51

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 18:41

breakubreaku

405

$endgroup$

share|cite|improve this question

edited Mar 13 at 18:51

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 18:41

breakubreaku

405

share|cite|improve this question

edited Mar 13 at 18:51

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 18:41

breakubreaku

405

edited Mar 13 at 18:51

Rodrigo de Azevedo

13.1k41960

edited Mar 13 at 18:51

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 18:41

breakubreaku

405

asked Mar 13 at 18:41

breakubreaku

405

1 Answer
1

active

oldest

votes

3

$begingroup$

$$y'' + 2y - 4 +frac1sqrty=0$$

$$2y''y'+4yy'-8y'+frac2y'sqrty=0$$
$$(y')^2+2y^2-8y+4sqrty=c_1$$
$y(0)=1$ and $y'(0)=2 quad;quad 4+2-8+4=c_1=2$
$$y'=pmsqrt-2y^2+8y-4sqrty+2$$
$$dx=pmfracdysqrt-2y^2+8y-4sqrty+2$$
$$x=pmint fracdysqrt-2y^2+8y-4sqrty+2+c_2$$
$y(0)=1$
$$x=pmint_1^y fracdxisqrt-2xi^2+8xi-4sqrtxi+2$$
$$x=pmint_1^sqrty frac2zeta:dzetasqrt-2zeta^4+8zeta^2-4zeta+2$$
This is the solution for the inverse function $x(y)$.

Further calculus appears arduous, involving elliptic integral and inverse.

Better use numerical calculus directly to solve the original ODE.

share|cite|improve this answer

answered Mar 14 at 16:03

JJacquelinJJacquelin

45k21855

$endgroup$

add a comment | 

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3

$begingroup$

$$y'' + 2y - 4 +frac1sqrty=0$$

$$2y''y'+4yy'-8y'+frac2y'sqrty=0$$
$$(y')^2+2y^2-8y+4sqrty=c_1$$
$y(0)=1$ and $y'(0)=2 quad;quad 4+2-8+4=c_1=2$
$$y'=pmsqrt-2y^2+8y-4sqrty+2$$
$$dx=pmfracdysqrt-2y^2+8y-4sqrty+2$$
$$x=pmint fracdysqrt-2y^2+8y-4sqrty+2+c_2$$
$y(0)=1$
$$x=pmint_1^y fracdxisqrt-2xi^2+8xi-4sqrtxi+2$$
$$x=pmint_1^sqrty frac2zeta:dzetasqrt-2zeta^4+8zeta^2-4zeta+2$$
This is the solution for the inverse function $x(y)$.

Further calculus appears arduous, involving elliptic integral and inverse.

Better use numerical calculus directly to solve the original ODE.

share|cite|improve this answer

answered Mar 14 at 16:03

JJacquelinJJacquelin

45k21855

$endgroup$

add a comment | 

3

$begingroup$

$$y'' + 2y - 4 +frac1sqrty=0$$

$$2y''y'+4yy'-8y'+frac2y'sqrty=0$$
$$(y')^2+2y^2-8y+4sqrty=c_1$$
$y(0)=1$ and $y'(0)=2 quad;quad 4+2-8+4=c_1=2$
$$y'=pmsqrt-2y^2+8y-4sqrty+2$$
$$dx=pmfracdysqrt-2y^2+8y-4sqrty+2$$
$$x=pmint fracdysqrt-2y^2+8y-4sqrty+2+c_2$$
$y(0)=1$
$$x=pmint_1^y fracdxisqrt-2xi^2+8xi-4sqrtxi+2$$
$$x=pmint_1^sqrty frac2zeta:dzetasqrt-2zeta^4+8zeta^2-4zeta+2$$
This is the solution for the inverse function $x(y)$.

Further calculus appears arduous, involving elliptic integral and inverse.

Better use numerical calculus directly to solve the original ODE.

share|cite|improve this answer

answered Mar 14 at 16:03

JJacquelinJJacquelin

45k21855

$endgroup$

add a comment | 

$begingroup$

$$y'' + 2y - 4 +frac1sqrty=0$$

$$2y''y'+4yy'-8y'+frac2y'sqrty=0$$
$$(y')^2+2y^2-8y+4sqrty=c_1$$
$y(0)=1$ and $y'(0)=2 quad;quad 4+2-8+4=c_1=2$
$$y'=pmsqrt-2y^2+8y-4sqrty+2$$
$$dx=pmfracdysqrt-2y^2+8y-4sqrty+2$$
$$x=pmint fracdysqrt-2y^2+8y-4sqrty+2+c_2$$
$y(0)=1$
$$x=pmint_1^y fracdxisqrt-2xi^2+8xi-4sqrtxi+2$$
$$x=pmint_1^sqrty frac2zeta:dzetasqrt-2zeta^4+8zeta^2-4zeta+2$$
This is the solution for the inverse function $x(y)$.

Further calculus appears arduous, involving elliptic integral and inverse.

Better use numerical calculus directly to solve the original ODE.

share|cite|improve this answer

answered Mar 14 at 16:03

JJacquelinJJacquelin

45k21855

$endgroup$

share|cite|improve this answer

answered Mar 14 at 16:03

JJacquelinJJacquelin

45k21855

answered Mar 14 at 16:03

JJacquelinJJacquelin

45k21855

answered Mar 14 at 16:03

JJacquelinJJacquelin

45k21855