Math

QuestionShow that the system y=Ay\mathbf{y}^{\prime}=A \mathbf{y} can be solved directly for a 2×22 \times 2 upper triangular matrix AA with constant entries. Use the form [xy]=[ab0c][xy]\left[\begin{array}{l} x \\ y \end{array}\right]^{\prime}=\left[\begin{array}{ll} a & b \\ 0 & c \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] and solve the second equation first.

Studdy Solution

STEP 1

Assumptions1. The system of equations is given by y=Ay\mathbf{y}^{\prime}=A \mathbf{y}, where AA is a × \times upper triangular matrix with constant entries. . The matrix AA is given by [ab0c]\left[\begin{array}{ll}a & b \\0 & c\end{array}\right], where a,b,cRa, b, c \in \mathbb{R}.
3. The vector y\mathbf{y} is given by [xy]\left[\begin{array}{l}x \\y\end{array}\right], where x,yx, y are functions of time.
4. The system can be solved directly by first solving the second equation, then the first.

STEP 2

First, let's write down the system of equations explicitly.
The first equation is given byx=ax+byx' = ax + byThe second equation is given byy=cyy' = cy

STEP 3

Let's solve the second equation first. This is a simple first order linear differential equation, which can be solved using the method of separation of variables.
y=cyy' = cy

STEP 4

Separate the variables and integrate both sides.
1ydy=cdt\int \frac{1}{y} dy = \int c dt

STEP 5

Evaluate the integrals.
lny=ct+C1\ln |y| = ct + C1

STEP 6

Exponentiate both sides to solve for yy.
y=ect+C1=eC1ect=C2ecty = e^{ct + C1} = e^{C1}e^{ct} = C2e^{ct}where C2=eC1C2 = e^{C1} is the new constant of integration.

STEP 7

Now, let's solve the first equation. This is a first order linear non-homogeneous differential equation, which can be solved using an integrating factor.
xax=byx' - ax = by

STEP 8

To find the integrating factor, we need to solve the homogeneous equation first.
xax=0x' - ax =0

STEP 9

Separate the variables and integrate both sides.
xdx=adt\int \frac{}{x} dx = \int a dt

STEP 10

Evaluate the integrals.
lnx=at+C3\ln |x| = at + C3

STEP 11

Exponentiate both sides to solve for xx.
x=eat+C3=eC3eat=C4eatx = e^{at + C3} = e^{C3}e^{at} = C4e^{at}where C4=eC3C4 = e^{C3} is the new constant of integration.

STEP 12

The integrating factor is eate^{at}.

STEP 13

Multiply the non-homogeneous equation by the integrating factor.
eatxaeatx=beatye^{at}x' - ae^{at}x = be^{at}y

STEP 14

This can be rewritten as a derivative of a product.
ddt(eatx)=beaty\frac{d}{dt}(e^{at}x) = be^{at}y

STEP 15

Integrate both sides with respect to tt.
d(eatx)=beatydt\int d(e^{at}x) = \int be^{at}y dt

STEP 16

Evaluate the integrals.
eatx=beatydt+C5e^{at}x = b\int e^{at}y dt + C5

STEP 17

Substitute y=C2ecty = C2e^{ct} into the integral.
eatx=beatC2ectdt+C5e^{at}x = b\int e^{at}C2e^{ct} dt + C5

STEP 18

Evaluate the integral.
eatx=bC2e(a+c)tdt+C5e^{at}x = bC2\int e^{(a+c)t} dt + C5

STEP 19

Evaluate the integral.
eatx=bCa+ce(a+c)t+C5e^{at}x = \frac{bC}{a+c}e^{(a+c)t} + C5

STEP 20

Finally, solve for xx by dividing both sides by eate^{at}.
x=bCa+cect+C5eatx = \frac{bC}{a+c}e^{ct} + C5e^{-at}The solution to the system of equations is given by[xy]=[bCa+cect+C5eatCect]\left[\begin{array}{l}x \\y\end{array}\right] = \left[\begin{array}{l}\frac{bC}{a+c}e^{ct} + C5e^{-at} \\Ce^{ct}\end{array}\right]

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord