Solving Differential Equations with Boundary Conditions

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:
\frac{dy}{dt}=ky
y(t)=y_0e^k^t
 
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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