Differential Equations: linearity principle

[Dusty912]

Homework Statement

Consider the linear system:
dx/dt=x-y
dy/dt=x+3y

a. show that the function (x(t), y(t))=(te^2t, -(t+1)e^2t) is a solution to the differential equation (easy)
b. Solve the initial value problem
dx/dt=x-y
dy/dt=x+3y

y(0)=(0,2)

need help with part b not a

Homework Equations

linarity principle
y(t)=k₁y₁(t) +k₂y₂(t)

The Attempt at a Solution

Really don't even know where to start. I understand how to use the linearity principle with two given solutions but not one. I feel like I am missing a very fundamental idea about initial value problems. So any idea on where I should begin with just one given solution?[/B]

 

[Mark44]

Homework Statement

Consider the linear system:
dx/dt=x-y
dy/dt=x+3y

a. show that the function (x(t), y(t))=(te^2t, -(t+1)e^2t) is a solution to the differential equation (easy)
b. Solve the initial value problem
dx/dt=x-y
dy/dt=x+3y

y(0)=(0,2)

This is confusing. y(0) is a single number. Did you write 0,2 to mean the same thing as 0.2? If not, what's the initial condition for x(0)?

need help with part b not a

Homework Equations

linarity principle

Which is what?

y(t)=k₁y₁(t) +k₂y₂(t)

The Attempt at a Solution

Really don't even know where to start. I understand how to use the linearity principle with two given solutions but not one. I feel like I am missing a very fundamental idea about initial value problems. So any idea on where I should begin with just one given solution?[/B]

 

[drvrm]

Homework Equations

p= Ɣmu
p= mv/ sqrt(1-v^2/c^2)

The Attempt at a Solution

a.
p= 1/ sqrt(1-.422c/c^2) *(.422c)(0)
= 5.51e-01

Really don't even know where to start. I understand how to use the linearity principle with two given solutions but not one. I feel like I am missing a very fundamental idea about initial value problems. So any idea on where I should begin with just one given solution?

please see the following to get an idea of the question-
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

 

[HallsofIvy]

Homework Statement

Consider the linear system:
dx/dt=x-y
dy/dt=x+3y

a. show that the function (x(t), y(t))=(te^2t, -(t+1)e^2t) is a solution to the differential equation (easy)
b. Solve the initial value problem
dx/dt=x-y
dy/dt=x+3y

y(0)=(0,2)

It looks like you are confusing two different notations. In your equations, dx/dt= x- y and dy/dt= x+ ,you have x and y as scalar functions but here you have y as a vector function with the previous x and y as components. In your original notation, you want x(0)= 0, y(0)= 2.

need help with part b not a

Homework Equations

linarity principle
y(t)=k₁y₁(t) +k₂y₂(t)

The Attempt at a Solution

Really don't even know where to start. I understand how to use the linearity principle with two given solutions but not one. I feel like I am missing a very fundamental idea about initial value problems. So any idea on where I should begin with just one given solution?[/B]

Are you required to use the given solution? There are several different ways of solving such a system of equations. For one, you could write it as a matrix equation:
$$\frac{d\begin{pmatrix}x \\ y \end{pmatrix}}{dx}= \begin{pmatrix}1 & -1 \\ 1 & 3 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}$$
and it is easy to "diagonalize" that matrix.

Or, differentiating the first equation again, $\frac{d^2x}{dt^2}= \frac{dx}{dt}- \frac{dy}{dt}$ and replace that $\frac{dy}{dx}$ with the second equation- $\frac{d^2x}{dt^2}= \frac{dx}{dt}- (x+ 3y)$ and replace that "y" with $y= x- \frac{dx}{dt}$ from the first equation again.
$\frac{d^2x}{dt^2}= \frac{dx}{dt}- x- 3(x- \frac{dx}{dt})= 4\frac{dx}{dt}- 4x$. Can you solve $\frac{d^2x}{dt^2}- 4\frac{dx}{dt}+ 4x= 0$ with initial conditions x(0)= (0), x'(0)= x(0)- y(0)= 0- 2= -2?

 

[Mark44]

It looks like you are confusing two different notations. In your equations, dx/dt= x- y and dy/dt= x+ ,you have x and y as scalar functions but here you have y as a vector function with the previous x and y as components. In your original notation, you want x(0)= 0, y(0)= 2.

Are you required to use the given solution? There are several different ways of solving such a system of equations. For one, you could write it as a matrix equation:
$$\frac{d\begin{pmatrix}x \\ y \end{pmatrix}}{dx}= \begin{pmatrix}1 & -1 \\ 1 & 3 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}$$
and it is easy to "diagonalize" that matrix.

It's not so easy. There's a repeated eigenvalue with only one eigenvector.

Or, differentiating the first equation again, $\frac{d^2x}{dt^2}= \frac{dx}{dt}- \frac{dy}{dt}$ and replace that $\frac{dy}{dx}$ with the second equation- $\frac{d^2x}{dt^2}= \frac{dx}{dt}- (x+ 3y)$ and replace that "y" with $y= x- \frac{dx}{dt}$ from the first equation again.
$\frac{d^2x}{dt^2}= \frac{dx}{dt}- x- 3(x- \frac{dx}{dt})= 4\frac{dx}{dt}- 4x$. Can you solve $\frac{d^2x}{dt^2}- 4\frac{dx}{dt}+ 4x= 0$ with initial conditions x(0)= (0), x'(0)= x(0)- y(0)= 0- 2= -2?