Undergrad Solve Linear System of ODEs: x',y' w/ Initial Conditions x(0),y(0)

[Euge]

Solve the linear system of ODE $x' = 2x + 3y$, $y' = -3x + y$ with initial conditions $x(0) = 1, y(0) = 2$.

 

[julian]

Spoiler

The differential equation

\begin{align*}
\frac{d}{dt} \binom{x}{y} =
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix}
\binom{x}{y}
\end{align*}

with initial conditions $x(0)=1$, $y(0) = 2$ has the solution:

\begin{align*}
\binom{x}{y} = \exp \left\{
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix} t \right\}
\binom{1}{2}
\end{align*}

When the eigenvalues are complex and distinct such that $\lambda_1 = \overline{\lambda_2} = a+ib$ we have that

\begin{align*}
e^{At} = e^{at} \cos (bt) \mathbb{1} + \dfrac{e^{at} \sin (bt)}{b} (A - a \mathbb{1}) \qquad (*) .
\end{align*}

We will prove this formula is correct by showing that $\frac{d}{dt} (RHS) = A (RHS)$ and by noting that the $RHS = \mathbb{1}$ at $t = 0$. By the Cayley-Hamilton theorem we have $A^2 = 2a A - (a^2+b^2) \mathbb{1}$. We use this in the form $A (A - a \mathbb{1}) = aA - (a^2 + b^2) \mathbb{1}$ in the proof of $\frac{d}{dt} (RHS) = A (RHS)$,

\begin{align*}
& \frac{d}{dt} \left[ e^{at} \cos (bt) \mathbb{1} + \dfrac{e^{at} \sin (bt)}{b} (A - a \mathbb{1}) \right]
\nonumber \\
& = a e^{at} \cos (bt) \mathbb{1} - b e^{at} \sin (bt) \mathbb{1} + \dfrac{a e^{at} \sin (bt)}{b} (A - a \mathbb{1}) + e^{at} \cos (bt) (A - a \mathbb{1})
\nonumber \\
& = A e^{at} \cos (bt) \mathbb{1} + \dfrac{e^{at} \sin (bt)}{b} (aA - (a^2 + b^2) \mathbb{1})
\nonumber \\
& = A \left[ e^{at} \cos (bt) \mathbb{1} + \dfrac{e^{at} \sin (bt)}{b} (A - a \mathbb{1}) \right] .
\end{align*}

The eigenvalues are determined by

\begin{align*}
\det
\begin{pmatrix}
2 - \lambda & 3 \\
-3 & 1 - \lambda
\end{pmatrix}
= 0
\end{align*}

or

\begin{align*}
\lambda^2 - 3 \lambda + 11 = 0
\end{align*}

so that

\begin{align*}
\lambda_1 = \frac{3}{2} + \frac{\sqrt{35}}{2} i , \quad \lambda_2 = \frac{3}{2} - \frac{\sqrt{35}}{2} i .
\end{align*}

So that by $(*)$ we have

\begin{align*}
e^{At} & = e^{3 t/2} \cos (\frac{ \sqrt{35} }{2} t)
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
+ e^{3 t/2} \dfrac{\sin (\frac{ \sqrt{35} }{2} t)}{\sqrt{35}}
\begin{pmatrix}
1 & 6 \\
-6 & -1
\end{pmatrix}
\end{align*}

So

\begin{align*}
\binom{x}{y} = e^{2 t/3} \cos (\frac{ \sqrt{35} }{2}t) \binom{1}{2}
+ e^{2 t/3} \dfrac{\sin (\frac{ \sqrt{35} }{2}t)}{\sqrt{35}} \binom{13}{-8}
\end{align*}

 

[julian]

In the previous post I wrote down slightly the wrong matrix. I've now written down the correct matrix and recalculated the eigenvalues.

By the way, there are other methods of solving the problem!

 

[erobz]

By the way, there are other methods of solving the problem!

Spoiler

The system is equivalent to two second order ODE's
$$ y'' - 3y'+11y , y(0)=2$$

$$ x''-3x'+11x=0, x(0)=1 $$

Algebraic error fixed ( hopefully)...

 

[Mayhem]

When working with these types of problems is it implied that x = x(t) and y = y(t).

 

[julian]

When working with these types of problems is it implied that x = x(t) and y = y(t).

Yes, they are functions of $t$.

 

[julian]

Spoiler

The system is equivalent to two second order ODE's
$$ y'' - 3y'+11y , y(0)=2$$

$$ x''-3x'+11x=0, x(0)=1 $$

Algebraic error fixed ( hopefully)...

Because you have a second order differential equation the general solution will involve two unknown constants, so you need $x(0)$ and $x'(0)$. Now, $x'(0)$ was not specified in the question but that does not present a problem because you are given a formula for $x'(t)$. Yep?

Do you know how to go about solving your differential equation? It is a homogeneous linear differential equation of second order with constant coefficients.

 

[erobz]

Because you have a second order differential equation the general solution will involve two unknown constants, so you need $x(0)$ and $x'(0)$. Now, $x'(0)$ was not specified in the question but that does not present a problem because you are given a formula for $x'(t)$. Yep?

Using the original equations and given initial conditions we find $x'(0) = 8, y'(0) = -1$

Do you know how to go about solving your differential equation? It is a homogeneous linear differential equation of second order with constant coefficients.

With an old textbook in front of me... Find the auxiliary equation:

$$m^2 -3m+11 = 0 \implies m = \frac{3}{2} \pm \frac{\sqrt{35}i}{2} $$

I think that implies the general solution ( for $y$ or $x$ - same ODE - different i.c. ):

$$ x(t) = y(t) = e^{ \left( \frac{3}{2}t \right) } \left( c_1 \cos \left( \frac{ \sqrt{35} }{2} t \right) + c_2 \sin \left( \frac{ \sqrt{35} }{2} t \right) \right) $$

 

[julian]

There are a couple of other methods of solving the problem.

Spoiler

Start with the "exponential ansatz":

\begin{align*}
\binom{x(t)}{y(t)} = \binom{C}{D} e^{\lambda t} .
\end{align*}

Substituting this into

\begin{align*}
\binom{x'}{y'} =
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix}
\binom{x}{y}
\end{align*}

we arrive at the eigenvector problem

\begin{align*}
\lambda \binom{C}{D} =
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix}
\binom{C}{D}
\end{align*}

Let's find the eigenvalues:

\begin{align*}
\det
\begin{pmatrix}
2 - \lambda & 3 \\
-3 & 1 - \lambda
\end{pmatrix}
= 0
\end{align*}

or

\begin{align*}
\lambda^2 - 3 \lambda + 11 = 0
\end{align*}

so that

\begin{align*}
\lambda_1 = \frac{3}{2} + i \frac{\sqrt{35}}{2} , \qquad \lambda_2 = \frac{3}{2} - i \frac{\sqrt{35}}{2}
\end{align*}

As the eigenvalues are the complex conjugate of each other one of the eigenvectors will be proportional to the complex conjugate of the other. The eigenvector equation is

\begin{align*}
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix}
\binom{C}{D} =
(\frac{3}{2} + i \frac{\sqrt{35}}{2}) \binom{C}{D}
\end{align*}

or

\begin{align*}
\begin{pmatrix}
4 & 6 \\
-6 & 2
\end{pmatrix}
\binom{C}{D} =
(3 + i \sqrt{35}) \binom{C}{D}
\end{align*}

or

\begin{align*}
\begin{pmatrix}
1 - i \sqrt{35} & 6 \\
-6 & -1 - i \sqrt{35}
\end{pmatrix}
\binom{C}{D} = 0
\end{align*}

or

\begin{align*}
(1 - i \sqrt{35}) C + 6 D = 0
\nonumber \\
-6 C - (1 + i \sqrt{35}) D = 0
\end{align*}

Using the first condition, the two eigenvectors are

\begin{align*}
\vec{v}_1 = \binom{1}{\dfrac{-1 + i \sqrt{35}}{6}} , \qquad (\lambda_1 = \frac{3}{2} + i \frac{\sqrt{35}}{2})
\nonumber \\
\vec{v}_2 = \binom{1}{\dfrac{-1 - i \sqrt{35}}{6}} , \qquad (\lambda_2 = \frac{3}{2} - i \frac{\sqrt{35}}{2})
\end{align*}

The (real) general solution of the homogeneous differential equation is

\begin{align*}
\binom{x}{y} = C \binom{1}{\dfrac{- 1 + i \sqrt{35}}{6}} e^{(\frac{3}{2} + i \frac{\sqrt{35}}{2}) t} + C^* \binom{1}{\dfrac{-1 - i \sqrt{35}}{6}} e^{(\frac{3}{2} - i \frac{\sqrt{35}}{2}) t}
\end{align*}

Writing $C = C_1 + i C_2$, we have

\begin{align*}
\binom{x}{y} = (C_1 + i C_2) \binom{1}{\dfrac{-1 + i \sqrt{35}}{6}} e^{(\frac{3}{2} + i \frac{\sqrt{35}}{2}) t} + c.c.
\end{align*}

so that

\begin{align*}
\binom{x}{y} = (C_1 + i C_2) \binom{1}{\dfrac{-1 + i \sqrt{35}}{6}} (\cos (\frac{\sqrt{35}}{2} t) + i \sin (\frac{\sqrt{35}}{2} t)) e^{3t/2} + c.c.
\end{align*}

or

\begin{align*}
\binom{x}{y} = \binom{C_1 + i C_2}{\dfrac{(-C_1 - C_2 \sqrt{35}) + i (-C_2 + C_1 \sqrt{35})}{6} } (\cos (\frac{\sqrt{35}}{2} t) + i \sin (\frac{\sqrt{35}}{2} t)) e^{3t/2} + c.c.
\end{align*}

or

\begin{align*}
\binom{x}{y} = 2 \binom{C_1 \cos (\frac{\sqrt{35}}{2} t) - C_2 \sin (\frac{\sqrt{35}}{2} t)}
{\dfrac{-C_1 - C_2 \sqrt{35}}{6} \cos (\frac{\sqrt{35}}{2} t) - \dfrac{-C_2 + C_1 \sqrt{35}}{6} \sin (\frac{\sqrt{35}}{2} t)} e^{3t/2}
\end{align*}

Using $x(0)=1$ and $y(0)=2$,

\begin{align*}
\binom{1}{2} = 2 \binom{ C_1 }{ - \dfrac{ C_1 + C_2 \sqrt{35} }{6} }
\end{align*}

So $C_1 = \frac{1}{2}$, $1 = - \dfrac{ C_1 + C_2 \sqrt{35} }{6}$, $C_2 = - \dfrac{13}{2 \sqrt{35}}$, and $\dfrac{4}{\sqrt{35}} = \dfrac{ -C_2 + C_1 \sqrt{35} }{6}$.

So finally we have,

\begin{align*}
\binom{x}{y} = e^{2 t/3} \cos (\frac{ \sqrt{35} }{2}t) \binom{1}{2}
+ e^{2 t/3} \dfrac{\sin (\frac{ \sqrt{35} }{2}t)}{\sqrt{35}} \binom{13}{-8} .
\end{align*}

Spoiler

The differential equation

\begin{align*}
\frac{d}{dt} \binom{x}{y} =
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix}
\binom{x}{y}
\end{align*}

with initial conditions $x(0)=1$, $y(0) = 2$ has the solution:

\begin{align*}
\binom{x}{y} = \exp \left\{
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix} t \right\}
\binom{1}{2}
\end{align*}

Write

\begin{align*}
A =
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix}
\end{align*}

We will compute $\exp (At)$.

To that end, consider the eigenvector problem

\begin{align*}
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix}
\binom{C}{D}
= \lambda \binom{C}{D}
\end{align*}

Let's find the eigenvalues:

\begin{align*}
\det
\begin{pmatrix}
2 - \lambda & 3 \\
-3 & 1 - \lambda
\end{pmatrix}
= 0
\end{align*}

or

\begin{align*}
\lambda^2 - 3 \lambda + 11 = 0
\end{align*}

so that

\begin{align*}
\lambda_1 = \frac{3}{2} + i \frac{\sqrt{35}}{2} , \qquad \lambda_2 = \frac{3}{2} - i \frac{\sqrt{35}}{2}
\end{align*}

As the eigenvalues are the complex conjugate of each other one of the eigenvectors will be proportional to the complex conjugate of the other. The eigenvector equation is

\begin{align*}
\begin{pmatrix}
2 & 3 \\
-3 & 1
\end{pmatrix}
\binom{C}{D} =
(\frac{3}{2} + i \frac{\sqrt{35}}{2}) \binom{C}{D}
\end{align*}

or

\begin{align*}
\begin{pmatrix}
4 & 6 \\
-6 & 2
\end{pmatrix}
\binom{C}{D} =
(3 + i \sqrt{35}) \binom{C}{D}
\end{align*}

or

\begin{align*}
\begin{pmatrix}
1 - i \sqrt{35} & 6 \\
-6 & -1 - i \sqrt{35}
\end{pmatrix}
\binom{C}{D} = 0
\end{align*}

or

\begin{align*}
(1 - i \sqrt{35}) C + 6 D = 0
\nonumber \\
-6 C - (1 + i \sqrt{35}) D = 0
\end{align*}

Using the first condition, the two normalised eigenvectors are

\begin{align*}
\vec{v}_1 = \frac{1}{\sqrt{7}} \binom{1}{\dfrac{-1 + i \sqrt{35}}{6}} , \qquad (\lambda_1 = \frac{3}{2} + i \frac{\sqrt{35}}{2})
\nonumber \\
\vec{v}_2 = \frac{1}{\sqrt{7}} \binom{1}{\dfrac{-1 - i \sqrt{35}}{6}} , \qquad (\lambda_2 = \frac{3}{2} - i \frac{\sqrt{35}}{2})
\end{align*}

Form the matrix $U$ whose columns are $\vec{v}_1$ and $\vec{v}_2$. Then

\begin{align*}
\exp ( A t ) & = U U^{-1} \exp ( A t ) U U^{-1}
\nonumber \\
& = U \left( \mathbb{1} + U^{-1} A U t + \frac{1}{2!} U^{-1} A U U^{-1} A U t^2 + \cdots \right) U^{-1}
\nonumber \\
& = U \exp ( \mathcal{D} t ) U^{-1}
\end{align*}

where

\begin{align*}
\mathcal{D} =
\begin{pmatrix}
\lambda_1 & 0 \\
0 & \lambda_2
\end{pmatrix}
=
\begin{pmatrix}
\frac{3}{2} + i \frac{\sqrt{35}}{2} & 0 \\
0 & \frac{3}{2} - i \frac{\sqrt{35}}{2}
\end{pmatrix} .
\end{align*}

We have

\begin{align*}
U = \frac{1}{6 \sqrt{7}}
\begin{pmatrix}
6 & 6 \\
-1+i\sqrt{35} & -1-i\sqrt{35}
\end{pmatrix}
\end{align*}

and

\begin{align*}
U^{-1} = \frac{\sqrt{35}}{10 \sqrt{7}}
\begin{pmatrix}
\sqrt{35}-i & -i6 \\
\sqrt{35}+i & i6
\end{pmatrix}
\end{align*}

and

\begin{align*}
\exp (\mathcal{D} t) = e^{3t/2}
\begin{pmatrix}
e^{i \sqrt{35}t/2} & 0 \\
0 & e^{-i \sqrt{35}t/2}
\end{pmatrix}
\end{align*}

We can now compute $\exp ( At )$,

\begin{align*}
& \exp (At)
\nonumber \\
& =
\frac{e^{3t/2} \sqrt{35}}{420}
\begin{pmatrix}
6 & 6 \\
-1+i\sqrt{35} & -1-i\sqrt{35}
\end{pmatrix}
\begin{pmatrix}
e^{i \sqrt{35}t/2} & 0 \\
0 & e^{-i \sqrt{35}t/2}
\end{pmatrix}
\begin{pmatrix}
\sqrt{35}-i & -i6 \\
\sqrt{35}+i & i6
\end{pmatrix}
\nonumber \\
& = \frac{e^{3t/2} \sqrt{35}}{420}
\begin{pmatrix}
6 & 6 \\
-1+i\sqrt{35} & -1-i\sqrt{35}
\end{pmatrix}
\begin{pmatrix}
(\sqrt{35}-i) e^{i \sqrt{35}t/2} & -i6 e^{i \sqrt{35}t/2} \\
(\sqrt{35}+i) e^{-i \sqrt{35}t/2} & i6 e^{-i \sqrt{35}t/2}
\end{pmatrix}
\nonumber \\
& = \frac{e^{\frac{3}{2}t} \sqrt{35}}{420}
\begin{pmatrix}
6 (\sqrt{35}-i) e^{i \frac{\sqrt{35}}{2}t} + 6 (\sqrt{35}+i) e^{-i \frac{\sqrt{35}}{2}t} & -36 i e^{i \frac{\sqrt{35}}{2}t} + i 36 e^{-i \frac{\sqrt{35}}{2}t} \\
36 i e^{i \frac{\sqrt{35}}{2}t} - i 36 e^{-i \frac{\sqrt{35}}{2}t} & i6 (1-i\sqrt{35}) e^{i \sqrt{35}t/2} - i6 (1+i\sqrt{35}) e^{-i \frac{\sqrt{35}}{2}t}
\end{pmatrix}
\nonumber \\
& = \frac{e^{\frac{3}{2}t} \sqrt{35}}{70}
\begin{pmatrix}
(\sqrt{35}-i) e^{i \frac{\sqrt{35}}{2}t} + (\sqrt{35}+i) e^{-i \frac{\sqrt{35}}{2}t} & -6 i e^{i \frac{\sqrt{35}}{2}t} + i 6 e^{-i \frac{\sqrt{35}}{2}t} \\
6 i e^{i \frac{\sqrt{35}}{2}t} - i 6 e^{-i \frac{\sqrt{35}}{2}t} & (i+\sqrt{35}) e^{i \frac{\sqrt{35}}{2}t} +(-i+\sqrt{35}) e^{-i \frac{\sqrt{35}}{2}t}
\end{pmatrix}
\nonumber \\
& = e^{3t/2}
\begin{pmatrix}
\cos (\sqrt{35}t/2) + \dfrac{\sin (\sqrt{35}t/2)}{\sqrt{35}} & 6 \dfrac{\sin (\sqrt{35}t/2)}{\sqrt{35}} \\
- 6 \dfrac{\sin (\sqrt{35}t/2)}{\sqrt{35}} & \cos (\sqrt{35}t/2) - \dfrac{\sin (\sqrt{35}t/2)}{\sqrt{35}}
\end{pmatrix}
\end{align*}

That is,

\begin{align*}
e^{At} & = e^{3 t/2} \cos (\frac{ \sqrt{35} }{2} t)
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
+ e^{3 t/2} \dfrac{\sin (\frac{ \sqrt{35} }{2} t)}{\sqrt{35}}
\begin{pmatrix}
1 & 6 \\
-6 & -1
\end{pmatrix}
\end{align*}

So

\begin{align*}
\binom{x}{y} = e^{2 t/3} \cos (\frac{ \sqrt{35} }{2}t) \binom{1}{2}
+ e^{2 t/3} \dfrac{\sin (\frac{ \sqrt{35} }{2}t)}{\sqrt{35}} \binom{13}{-8}
\end{align*}

 

[bob012345]

Using the original equations and given initial conditions we find $x'(0) = 8, y'(0) = -1$
With an old textbook in front of me... Find the auxiliary equation:

$$m^2 -3m+11 = 0 \implies m = \frac{3}{2} \pm \frac{\sqrt{35}i}{2} $$

I think that implies the general solution ( for $y$ or $x$ - same ODE - different i.c. ):

$$ x(t) = y(t) = e^{ \left( \frac{3}{2}t \right) } \left( c_1 \cos \left( \frac{ \sqrt{35} }{2} t \right) + c_2 \sin \left( \frac{ \sqrt{35} }{2} t \right) \right) $$

But different constants in each case, not the same.

 

[erobz]

But different constants in each case, not the same.

Yeah, I was just being lazy. Thats why I said the general solution is... the constants would be determined for the initial conditions of each. I would think those constants aren't necessarily distinct in the general solution...so I thought I could get away with it?