Differential Equations

Thread Starter

Vikram50517

Joined Jan 4, 2020
81
Consider a differential equation dy/dt +(t^2)y=0. This equation is said to be a linear DE in y.Does it mean that when dy/dx is plotted in y axis and y in x axis(any one of the solutions),would the graph be a straight line?,and the next thing (t^2) is a non linear term and t^2 can be expressed as a inverse function of y which is a non linear function in y ,which when substituted in the DE obviously makes the equation non linear,but the DE is a linear equation in y ,how is that possible?
 

Papabravo

Joined Feb 24, 2006
21,159
Consider a differential equation dy/dt +(t^2)y=0. This equation is said to be a linear DE in y.Does it mean that when dy/dx is plotted in y axis and y in x axis(any one of the solutions),would the graph be a straight line?,and the next thing (t^2) is a non linear term and t^2 can be expressed as a inverse function of y which is a non linear function in y ,which when substituted in the DE obviously makes the equation non linear,but the DE is a linear equation in y ,how is that possible?
It is not a linear differential equation, as I understand the term. In order for it to be true and have a solution the derivative has to tend towards -∞ as t^2 approaches +∞ at the same rate. A feature of well behaved differential equations is that bounded inputs have bounded outputs.
 

MrAl

Joined Jun 17, 2014
11,389
It is not a linear differential equation, as I understand the term. In order for it to be true and have a solution the derivative has to tend towards -∞ as t^2 approaches +∞ at the same rate. A feature of well behaved differential equations is that bounded inputs have bounded outputs.
The way i understand it is that the equation given is LINEAR.
That is because any variable with 'y' is either just a derivative or a first power of y possibly with a constant coefficient.
Examples (y'=dy/dt):
y"+2*y'+y=0 LINEAR
y*y'+2*y=0 NONLINEAR
y'+(t^3+t)*y=0 LINEAR
y'+y^2=0 NONLINEAR
y'+sqrt(y)=0 NONLINAR
y'+sqrt(t)=0 LINEAR

So the main rule is that the independent variable can be any function but the dependent variable has to be first degree or a lone derivative, and can not be multiplied or combined with another derivative or function of the dependent variable.
So anything with a function like t^2, t^3, sin(t) is linear, but
anything with y^2, sin(y), y'*y", y*y', sqrt(y), is non linear.
 

bogosort

Joined Sep 24, 2011
696
Consider a differential equation dy/dt +(t^2)y=0. This equation is said to be a linear DE in y.Does it mean that when dy/dx is plotted in y axis and y in x axis(any one of the solutions),would the graph be a straight line?
No, the solutions of this DE are decaying exponentials, which are clearly not straight lines. In this context, linearity refers to the more general concept of linear operators.

Every homogeneous linear DE (like the one in your example) can be written in operator form as \( Ly = 0 \), where \( L \) is a linear operator satisfying the linearity conditions:
\[ L( c y_1 + c y_2 ) = c L(y_1) + c L(y_2) \]
where \( y_1 \) and \( y_2 \) are differentiable functions, and \( c \) a scalar constant.

In your example, \( L = (\frac{d}{dt} + t^2) \) acts on the subspace of once-differentiable functions, and the solutions of your DE are the kernel of \( L \), i.e., those functions y for which \( Ly = 0 \).

A differential equation is nonlinear when its operator fails to satisfy the linearity conditions, which has nothing to do with the shape of the graphs of the solutions.

and the next thing (t^2) is a non linear term and t^2 can be expressed as a inverse function of y which is a non linear function in y ,which when substituted in the DE obviously makes the equation non linear,but the DE is a linear equation in y ,how is that possible?
Since the operator in your equation acts on y, not t, it doesn't matter what t is doing. You can think of the \( t^2 \) factor as a non-constant coefficient:
\[ \frac{dy}{dt} = a(t) y(t) \]
 

Thread Starter

Vikram50517

Joined Jan 4, 2020
81
It is not a linear differential equation, as I understand the term. In order for it to be true and have a solution the derivative has to tend towards -∞ as t^2 approaches +∞ at the same rate. A feature of well behaved differential equations is that bounded inputs have bounded outputs.
Thank you very much
 

Thread Starter

Vikram50517

Joined Jan 4, 2020
81
It is not a linear differential equation, as I understand the term. In order for it to be true and have a solution the derivative has to tend towards -∞ as t^2 approaches +∞ at the same rate. A feature of well behaved differential equations is that bounded inputs have bounded outputs.
Thank you very much
The way i understand it is that the equation given is LINEAR.
That is because any variable with 'y' is either just a derivative or a first power of y possibly with a constant coefficient.
Examples (y'=dy/dt):
y"+2*y'+y=0 LINEAR
y*y'+2*y=0 NONLINEAR
y'+(t^3+t)*y=0 LINEAR
y'+y^2=0 NONLINEAR
y'+sqrt(y)=0 NONLINAR
y'+sqrt(t)=0 LINEAR

So the main rule is that the independent variable can be any function but the dependent variable has to be first degree or a lone derivative, and can not be multiplied or combined with another derivative or function of the dependent variable.
So anything with a function like t^2, t^3, sin(t) is linear, but
anything with y^2, sin(y), y'*y", y*y', sqrt(y), is non linear.
Thank you very much
 

Thread Starter

Vikram50517

Joined Jan 4, 2020
81
No, the solutions of this DE are decaying exponentials, which are clearly not straight lines. In this context, linearity refers to the more general concept of linear operators.

Every homogeneous linear DE (like the one in your example) can be written in operator form as \( Ly = 0 \), where \( L \) is a linear operator satisfying the linearity conditions:
\[ L( c y_1 + c y_2 ) = c L(y_1) + c L(y_2) \]
where \( y_1 \) and \( y_2 \) are differentiable functions, and \( c \) a scalar constant.

In your example, \( L = (\frac{d}{dt} + t^2) \) acts on the subspace of once-differentiable functions, and the solutions of your DE are the kernel of \( L \), i.e., those functions y for which \( Ly = 0 \).

A differential equation is nonlinear when its operator fails to satisfy the linearity conditions, which has nothing to do with the shape of the graphs of the solutions.


Since the operator in your equation acts on y, not t, it doesn't matter what t is doing. You can think of the \( t^2 \) factor as a non-constant coefficient:
\[ \frac{dy}{dt} = a(t) y(t) \]
So dy/dt =a(t)*y(t) is just a differential equation with variable coefficients such that the coefficients does not affect the linearity of the equation in any way i.e the mapping process is independent of the coefficients am I right?
 

Thread Starter

Vikram50517

Joined Jan 4, 2020
81
No, the solutions of this DE are decaying exponentials, which are clearly not straight lines. In this context, linearity refers to the more general concept of linear operators.

Every homogeneous linear DE (like the one in your example) can be written in operator form as \( Ly = 0 \), where \( L \) is a linear operator satisfying the linearity conditions:
\[ L( c y_1 + c y_2 ) = c L(y_1) + c L(y_2) \]
where \( y_1 \) and \( y_2 \) are differentiable functions, and \( c \) a scalar constant.

In your example, \( L = (\frac{d}{dt} + t^2) \) acts on the subspace of once-differentiable functions, and the solutions of your DE are the kernel of \( L \), i.e., those functions y for which \( Ly = 0 \).

A differential equation is nonlinear when its operator fails to satisfy the linearity conditions, which has nothing to do with the shape of the graphs of the solutions.


Since the operator in your equation acts on y, not t, it doesn't matter what t is doing. You can think of the \( t^2 \) factor as a non-constant coefficient:
\[ \frac{dy}{dt} = a(t) y(t) \]
 

Thread Starter

Vikram50517

Joined Jan 4, 2020
81
No, the solutions of this DE are decaying exponentials, which are clearly not straight lines. In this context, linearity refers to the more general concept of linear operators.

Every homogeneous linear DE (like the one in your example) can be written in operator form as \( Ly = 0 \), where \( L \) is a linear operator satisfying the linearity conditions:
\[ L( c y_1 + c y_2 ) = c L(y_1) + c L(y_2) \]
where \( y_1 \) and \( y_2 \) are differentiable functions, and \( c \) a scalar constant.

In your example, \( L = (\frac{d}{dt} + t^2) \) acts on the subspace of once-differentiable functions, and the solutions of your DE are the kernel of \( L \), i.e., those functions y for which \( Ly = 0 \).

A differential equation is nonlinear when its operator fails to satisfy the linearity conditions, which has nothing to do with the shape of the graphs of the solutions.


Since the operator in your equation acts on y, not t, it doesn't matter what t is doing. You can think of the \( t^2 \) factor as a non-constant coefficient:
\[ \frac{dy}{dt} = a(t) y(t) \]
As you said (d/dt + t^2) is a linear operator acting on the mathematical space y.so then it maps the input which is y to the function (d/dt + t^2)y
Which is obviously a linear mapping because the operator is a linear operator.Then if I plot (d/dt + t^2)y vs y ( take any particular exponential solution :) )how will the plot be?
 

bogosort

Joined Sep 24, 2011
696
As you said (d/dt + t^2) is a linear operator acting on the mathematical space y.so then it maps the input which is y to the function (d/dt + t^2)y
Which is obviously a linear mapping because the operator is a linear operator.Then if I plot (d/dt + t^2)y vs y ( take any particular exponential solution :) )how will the plot be?
The solutions of this DE are the family of functions represented by \( y = C e^{-t^3 / 3} \). So, the plot of any particular solution is a growing exponential as \( t \to -\infty \) and a decaying exponential as \( t \to +\infty \), which join at t = 0 at y = C.

If you're looking for intuition, think of a DE as describing the time-evolving dynamics of a physical system. In your example, y(t) is the output of the system at time t. As your DE is homogeneous (equal to 0), there is no input to the system, thus the solutions characterize the transient response of the system, i.e., the dynamics of the system itself. In particular, your DE tells us that the system is stable for any causal input (x(t) = 0 for all t < 0).
 
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