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Definition linear ODE

observing linearity in differential equationsGeneral solution to 3D linear 2nd order PDE using Wronskian and Integrals?Determine the properties of the solution of a second order linear ODESome clarification on this textbook's definition of linear ODEs?when two solutions do not form a basis for the solution space to a 2nd order odeGeneral solution of this ODEFormula to find a particular solution of a specific linear ODEReduction of Order on Second Order Linear Homogenous ODE ExplanationDefinition of ordinary derivativeRiccati ODE Solution Path

2

$begingroup$

An ordinary differential equation is said to be linear if
$$F(t,y(t),...,y^{(n)}(t))=0$$
is linear in every derivative. I run into a little problem when using this definition for the equation
$$y'y=0$$

because we have that $$F(t,alpha y_1(t)+beta y_2(t),y'(t))=(alpha y_1(t)+beta y_2(t))y'(t)=alpha y_1(t)y't(t) + beta y_2(t)y'(t)\
=alpha F(t,y_1(t),y'(t)) + beta F(t,y_2(t),y'(t))$$

and vice versa for $y'(t)$. This shows that the ODE is linear. But an equivalent definition of linearity states that it has to have the form

$F(t,y(t),...,y^{(n)}(t))=sum_{k=0}^n a_i(t)y^{(i)}(t) - g(t)=0$

With this definition the ODE is not linear anymore. Not sure what my mistake is there.

ordinary-differential-equations

share|cite|improve this question

asked Mar 19 at 16:26

EpsilonDeltaEpsilonDelta

7301615

$endgroup$

add a comment | 

2

$begingroup$

An ordinary differential equation is said to be linear if
$$F(t,y(t),...,y^{(n)}(t))=0$$
is linear in every derivative. I run into a little problem when using this definition for the equation
$$y'y=0$$

because we have that $$F(t,alpha y_1(t)+beta y_2(t),y'(t))=(alpha y_1(t)+beta y_2(t))y'(t)=alpha y_1(t)y't(t) + beta y_2(t)y'(t)\
=alpha F(t,y_1(t),y'(t)) + beta F(t,y_2(t),y'(t))$$

and vice versa for $y'(t)$. This shows that the ODE is linear. But an equivalent definition of linearity states that it has to have the form

$F(t,y(t),...,y^{(n)}(t))=sum_{k=0}^n a_i(t)y^{(i)}(t) - g(t)=0$

With this definition the ODE is not linear anymore. Not sure what my mistake is there.

ordinary-differential-equations

share|cite|improve this question

asked Mar 19 at 16:26

EpsilonDeltaEpsilonDelta

7301615

$endgroup$

add a comment | 

$begingroup$

An ordinary differential equation is said to be linear if
$$F(t,y(t),...,y^{(n)}(t))=0$$
is linear in every derivative. I run into a little problem when using this definition for the equation
$$y'y=0$$

because we have that $$F(t,alpha y_1(t)+beta y_2(t),y'(t))=(alpha y_1(t)+beta y_2(t))y'(t)=alpha y_1(t)y't(t) + beta y_2(t)y'(t)\
=alpha F(t,y_1(t),y'(t)) + beta F(t,y_2(t),y'(t))$$

and vice versa for $y'(t)$. This shows that the ODE is linear. But an equivalent definition of linearity states that it has to have the form

$F(t,y(t),...,y^{(n)}(t))=sum_{k=0}^n a_i(t)y^{(i)}(t) - g(t)=0$

With this definition the ODE is not linear anymore. Not sure what my mistake is there.

ordinary-differential-equations

share|cite|improve this question

asked Mar 19 at 16:26

EpsilonDeltaEpsilonDelta

7301615

$endgroup$

share|cite|improve this question

asked Mar 19 at 16:26

EpsilonDeltaEpsilonDelta

7301615

share|cite|improve this question

asked Mar 19 at 16:26

EpsilonDeltaEpsilonDelta

7301615

asked Mar 19 at 16:26

EpsilonDeltaEpsilonDelta

7301615

asked Mar 19 at 16:26

EpsilonDeltaEpsilonDelta

7301615

2 Answers
2

active

oldest

votes

2

$begingroup$

The problem is $y^{(n)}$ are dependent on $y$, so your definition is not sound. Derivative operators are linear by default, so the equation is linear if it is linear in $y$

Let

$$ F(y) = y'y $$

Then

$$ F(ay_1 + by_2) = (ay_1' + by_2')(ay_1+by_2) $$

while

$$ aF(y_1) + bF(y_2) = ay_1'y_1 + by_2'y_2 $$

which means

$$ F(ay_1+by_2) ne aF(y_1) + bF(y_2) $$

Therefore $F(y)$ is not a linear operator.

share|cite|improve this answer

answered Mar 19 at 16:52

DylanDylan

14.2k31127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Finally, thanks. Can you elaborate on what you mean by "not sound". Do you mean misleading, since all derivatives are dependend on the function itself?
  $endgroup$
  – EpsilonDelta
  Mar 19 at 17:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  By that, I mean the way that you applied it is not correct. $F(t,ay_1+by_2,y')$ doesn't make sense, since $y'$ cannot be independent of $y$.
  $endgroup$
  – Dylan
  Mar 19 at 17:15
  

  
  
  
  

add a comment | 

3

$begingroup$

We have $F(t,y(t),y’(t))=yy’$. Note that $$F(alpha t, alpha y(t), alpha y’(t)) =alpha^2yy’ neq alpha yy’=alpha F(t,y(t),y’(t))$$

and thus $F$ is not linear in its arguments, implying that it is not a linear differential equation using your first definition.

share|cite|improve this answer

answered Mar 19 at 16:45

Tyler6Tyler6

736414

$endgroup$

add a comment | 

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2

$begingroup$

The problem is $y^{(n)}$ are dependent on $y$, so your definition is not sound. Derivative operators are linear by default, so the equation is linear if it is linear in $y$

Let

$$ F(y) = y'y $$

Then

$$ F(ay_1 + by_2) = (ay_1' + by_2')(ay_1+by_2) $$

while

$$ aF(y_1) + bF(y_2) = ay_1'y_1 + by_2'y_2 $$

which means

$$ F(ay_1+by_2) ne aF(y_1) + bF(y_2) $$

Therefore $F(y)$ is not a linear operator.

share|cite|improve this answer

answered Mar 19 at 16:52

DylanDylan

14.2k31127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Finally, thanks. Can you elaborate on what you mean by "not sound". Do you mean misleading, since all derivatives are dependend on the function itself?
  $endgroup$
  – EpsilonDelta
  Mar 19 at 17:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  By that, I mean the way that you applied it is not correct. $F(t,ay_1+by_2,y')$ doesn't make sense, since $y'$ cannot be independent of $y$.
  $endgroup$
  – Dylan
  Mar 19 at 17:15
  

  
  
  
  

add a comment | 

2

$begingroup$

The problem is $y^{(n)}$ are dependent on $y$, so your definition is not sound. Derivative operators are linear by default, so the equation is linear if it is linear in $y$

Let

$$ F(y) = y'y $$

Then

$$ F(ay_1 + by_2) = (ay_1' + by_2')(ay_1+by_2) $$

while

$$ aF(y_1) + bF(y_2) = ay_1'y_1 + by_2'y_2 $$

which means

$$ F(ay_1+by_2) ne aF(y_1) + bF(y_2) $$

Therefore $F(y)$ is not a linear operator.

share|cite|improve this answer

answered Mar 19 at 16:52

DylanDylan

14.2k31127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Finally, thanks. Can you elaborate on what you mean by "not sound". Do you mean misleading, since all derivatives are dependend on the function itself?
  $endgroup$
  – EpsilonDelta
  Mar 19 at 17:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  By that, I mean the way that you applied it is not correct. $F(t,ay_1+by_2,y')$ doesn't make sense, since $y'$ cannot be independent of $y$.
  $endgroup$
  – Dylan
  Mar 19 at 17:15
  

  
  
  
  

add a comment | 

$begingroup$

The problem is $y^{(n)}$ are dependent on $y$, so your definition is not sound. Derivative operators are linear by default, so the equation is linear if it is linear in $y$

Let

$$ F(y) = y'y $$

Then

$$ F(ay_1 + by_2) = (ay_1' + by_2')(ay_1+by_2) $$

while

$$ aF(y_1) + bF(y_2) = ay_1'y_1 + by_2'y_2 $$

which means

$$ F(ay_1+by_2) ne aF(y_1) + bF(y_2) $$

Therefore $F(y)$ is not a linear operator.

share|cite|improve this answer

answered Mar 19 at 16:52

DylanDylan

14.2k31127

$endgroup$

share|cite|improve this answer

answered Mar 19 at 16:52

DylanDylan

14.2k31127

answered Mar 19 at 16:52

DylanDylan

14.2k31127

answered Mar 19 at 16:52

DylanDylan

14.2k31127

3

$begingroup$

We have $F(t,y(t),y’(t))=yy’$. Note that $$F(alpha t, alpha y(t), alpha y’(t)) =alpha^2yy’ neq alpha yy’=alpha F(t,y(t),y’(t))$$

and thus $F$ is not linear in its arguments, implying that it is not a linear differential equation using your first definition.

share|cite|improve this answer

answered Mar 19 at 16:45

Tyler6Tyler6

736414

$endgroup$

add a comment | 

3

$begingroup$

We have $F(t,y(t),y’(t))=yy’$. Note that $$F(alpha t, alpha y(t), alpha y’(t)) =alpha^2yy’ neq alpha yy’=alpha F(t,y(t),y’(t))$$

and thus $F$ is not linear in its arguments, implying that it is not a linear differential equation using your first definition.

share|cite|improve this answer

answered Mar 19 at 16:45

Tyler6Tyler6

736414

$endgroup$

add a comment | 

$begingroup$

We have $F(t,y(t),y’(t))=yy’$. Note that $$F(alpha t, alpha y(t), alpha y’(t)) =alpha^2yy’ neq alpha yy’=alpha F(t,y(t),y’(t))$$

and thus $F$ is not linear in its arguments, implying that it is not a linear differential equation using your first definition.

share|cite|improve this answer

answered Mar 19 at 16:45

Tyler6Tyler6

736414

$endgroup$

share|cite|improve this answer

answered Mar 19 at 16:45

Tyler6Tyler6

736414

answered Mar 19 at 16:45

Tyler6Tyler6

736414

answered Mar 19 at 16:45

Tyler6Tyler6

736414