System of two differential equations

[docnet]

Homework Statement

Solve the system of differential equations

Relevant Equations

x''-3x'+2x = 0 , x(0)= u
y'+y^2cot(t + pi/2)=0

The first equation leads to x = ae^2t + be^t
and the second equation leads to y=[1/(ln(sint+pi/2)+c)]

this corresponds to the system

a+b=1/c
2a+b=1

which has infinitely many solutions. what am I missing here?

 

[fresh_42]

Homework Statement:: Solve the system of differential equations
Relevant Equations:: x''-3x'+2x = 0 , x(0)= u
y'+y^2cot(t + pi/2)=0

View attachment 271647

The first equation leads to x = ae^2t + be^t
and the second equation leads to y=[1/(ln(sint+pi/2)+c)]

this corresponds to the system

a+b=1/c
2a+b=1

which has infinitely many solutions. what am I missing here?

You have no initial condition for $y(0)$ so you get a set of solutions. All you can do is express $a,b$ in terms of $c$.

 

[docnet]

Okay, then

a = 1-(1/c)
b = (2/c) - 1

and the solution is

x = ((1-(1/c))e^2t + ((2/c)-1)e^t
y = 1/[ln(sint+pi/2)+c]

 

[fresh_42]

At least I don't see what else could be done without additional information. The second equation is not defined at $t=0$ but that doesn't mean that $y$ isn't defined. But we do not know $\dot{y}(0)$, so we cannot perform any limit considerations.

 

[haruspex]

Doesn't look like a system of equations to me. It's two independent equations sharing a common initial value. I strongly suspect a misprint.

 

[Delta2]

Doesn't look like a system of equations to me. It's two independent equations sharing a common initial value. I strongly suspect a misprint.

Thats what i thought at my first glance too, but then i realized that the coupling is in the initial condition, which creates a coupling in the constants that appear in the two solutions.

 

[haruspex]

Thats what i thought at my first glance too, but then i realized that the coupling is in the initial condition, which creates a coupling in the constants that appear in the two solutions.

Sure, but the process of solving each is completely independent of the other. More socially distanced than coupled.

 

[Delta2]

Sure, but the process of solving each is completely independent of the other. More socially distanced than coupled.

Well i kind of agree, the processes of solving are independent (up to the constants) however the two solutions are coupled because the constants that appear in them are coupled.