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Show that satisfy the ODE

0

$begingroup$

Let be $f(x)$ a function defined for all real numbers. Suppose that $cos(x)$ is solution to the ODE: $y'=f(y)$. Show that $-sin(x)$ also is a solution.

I tried using the fact that is a separable ODE, then I have

$$int_{x_0}^{cos(x)} frac{ds}{f(s)} = x + c_1$$ that implies

$$int_{cos(x_0)}^{x} frac{-sin(s)}{f(cos(s))} = x + c_1$$ and If I show that $$int frac{-cos(x)}{f(-sin(x))} = x + c_1$$ then I show that $-sin(x)$ is a solution?

Thank you.

ordinary-differential-equations

share|cite|improve this question

edited Mar 24 at 0:36

Rodrigo Palacios

asked Mar 23 at 23:39

Rodrigo PalaciosRodrigo Palacios

235

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What does sen mean?
  $endgroup$
  – Paul
  Mar 23 at 23:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Paul Sorry, is sin(x)
  $endgroup$
  – Rodrigo Palacios
  Mar 23 at 23:49
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the differential in the second equation?
  $endgroup$
  – herb steinberg
  Mar 23 at 23:52
  

  
  
  
  

add a comment | 

0

$begingroup$

Let be $f(x)$ a function defined for all real numbers. Suppose that $cos(x)$ is solution to the ODE: $y'=f(y)$. Show that $-sin(x)$ also is a solution.

I tried using the fact that is a separable ODE, then I have

$$int_{x_0}^{cos(x)} frac{ds}{f(s)} = x + c_1$$ that implies

$$int_{cos(x_0)}^{x} frac{-sin(s)}{f(cos(s))} = x + c_1$$ and If I show that $$int frac{-cos(x)}{f(-sin(x))} = x + c_1$$ then I show that $-sin(x)$ is a solution?

Thank you.

ordinary-differential-equations

share|cite|improve this question

edited Mar 24 at 0:36

Rodrigo Palacios

asked Mar 23 at 23:39

Rodrigo PalaciosRodrigo Palacios

235

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What does sen mean?
  $endgroup$
  – Paul
  Mar 23 at 23:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Paul Sorry, is sin(x)
  $endgroup$
  – Rodrigo Palacios
  Mar 23 at 23:49
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the differential in the second equation?
  $endgroup$
  – herb steinberg
  Mar 23 at 23:52
  

  
  
  
  

add a comment | 

$begingroup$

Let be $f(x)$ a function defined for all real numbers. Suppose that $cos(x)$ is solution to the ODE: $y'=f(y)$. Show that $-sin(x)$ also is a solution.

I tried using the fact that is a separable ODE, then I have

$$int_{x_0}^{cos(x)} frac{ds}{f(s)} = x + c_1$$ that implies

$$int_{cos(x_0)}^{x} frac{-sin(s)}{f(cos(s))} = x + c_1$$ and If I show that $$int frac{-cos(x)}{f(-sin(x))} = x + c_1$$ then I show that $-sin(x)$ is a solution?

Thank you.

ordinary-differential-equations

share|cite|improve this question

edited Mar 24 at 0:36

Rodrigo Palacios

asked Mar 23 at 23:39

Rodrigo PalaciosRodrigo Palacios

235

$endgroup$

share|cite|improve this question

edited Mar 24 at 0:36

Rodrigo Palacios

asked Mar 23 at 23:39

Rodrigo PalaciosRodrigo Palacios

235

share|cite|improve this question

edited Mar 24 at 0:36

Rodrigo Palacios

asked Mar 23 at 23:39

Rodrigo PalaciosRodrigo Palacios

235

asked Mar 23 at 23:39

Rodrigo PalaciosRodrigo Palacios

235

asked Mar 23 at 23:39

Rodrigo PalaciosRodrigo Palacios

235

1 Answer
1

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oldest

votes

3

$begingroup$

You could observe that
$$-sin(x) = cosleft(frac{pi}{2}+xright)$$
and that if
$$y(x) = cos(x)$$
then $y'(x) =-sin(x) = f(cos(x))$
and so we can say that
$$y(x+pi/2) = cos(x+pi/2)$$
which implies that
$$y'(x+pi/2) =-sin(x+pi/2) = f(cos(x+pi/2)).$$

share|cite|improve this answer

answered Mar 23 at 23:51

model_checkermodel_checker

4,45521931

$endgroup$

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3

$begingroup$

You could observe that
$$-sin(x) = cosleft(frac{pi}{2}+xright)$$
and that if
$$y(x) = cos(x)$$
then $y'(x) =-sin(x) = f(cos(x))$
and so we can say that
$$y(x+pi/2) = cos(x+pi/2)$$
which implies that
$$y'(x+pi/2) =-sin(x+pi/2) = f(cos(x+pi/2)).$$

share|cite|improve this answer

answered Mar 23 at 23:51

model_checkermodel_checker

4,45521931

$endgroup$

add a comment | 

3

$begingroup$

You could observe that
$$-sin(x) = cosleft(frac{pi}{2}+xright)$$
and that if
$$y(x) = cos(x)$$
then $y'(x) =-sin(x) = f(cos(x))$
and so we can say that
$$y(x+pi/2) = cos(x+pi/2)$$
which implies that
$$y'(x+pi/2) =-sin(x+pi/2) = f(cos(x+pi/2)).$$

share|cite|improve this answer

answered Mar 23 at 23:51

model_checkermodel_checker

4,45521931

$endgroup$

add a comment | 

$begingroup$

You could observe that
$$-sin(x) = cosleft(frac{pi}{2}+xright)$$
and that if
$$y(x) = cos(x)$$
then $y'(x) =-sin(x) = f(cos(x))$
and so we can say that
$$y(x+pi/2) = cos(x+pi/2)$$
which implies that
$$y'(x+pi/2) =-sin(x+pi/2) = f(cos(x+pi/2)).$$

share|cite|improve this answer

answered Mar 23 at 23:51

model_checkermodel_checker

4,45521931

$endgroup$

share|cite|improve this answer

answered Mar 23 at 23:51

model_checkermodel_checker

4,45521931

answered Mar 23 at 23:51

model_checkermodel_checker

4,45521931

answered Mar 23 at 23:51

model_checkermodel_checker

4,45521931