Solving Differential Equations with Boundary Conditions

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SUMMARY

The discussion centers on solving a system of differential equations defined by dx/dt = -k1 (y) and dy/dt = k2 x, with constants k1 and k2 both set to 10, and initial conditions x(0) = 100 and y(0) = 0. The proposed solution involves substituting variables leading to a second-order differential equation d2y/dt2 = -100y, resulting in the general solution Y = c1 cos(10x) + c2 sin(10x). However, participants point out the incorrect introduction of 'x' in the solution, emphasizing the need to re-evaluate the approach to applying boundary conditions correctly.

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  • Understanding of first-order and second-order differential equations
  • Familiarity with boundary value problems in differential equations
  • Knowledge of initial conditions and their application in solving equations
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  • Review methods for solving second-order linear differential equations
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Students, mathematicians, and engineers who are working with differential equations, particularly those focusing on boundary value problems and initial condition applications.

supermesh
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We have these two equations:
dx/dt = -k1 (y)
dy/dt = k2 x
k1= 10 k2= 10
These are the conditions:
x(0)= 100
y(0) = 0

I am so confused...This is what I have done so far...but is not making any sense...

x= (dy/dt) (1/k2)
Substitute in:

d2y/dt2 = -100 y

So the solution is

Y= c1 cos 10x+ c2 sin 10x

I don't know how to use boundary counditions now...Is this even a right approach?

Thank you so much
 
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The rule for a differential equation:
[tex]\frac{dy}{dt}=ky[/tex]
[tex]y(t)=y_0e^k^t[/tex]
 
supermesh said:
x= (dy/dt) (1/k2)
Substitute in:

d2y/dt2 = -100 y
Good so far.

So the solution is

Y= c1 cos 10x+ c2 sin 10x
There is no x in your differential equation; why do you have one here? Try finding the solution to the second order differential equation again.
 

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