An Interesting System of Differential Equations
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I enjoyed watching this because it's describing the work of a matrix. The derivative of a vector d/dt() is equal to a 2x2 matrix A multiplied by added to the vector . The 2x2 matrix A has entries [1, 1 \\ 1, -1]. The eigenvalues of this matrix are sqrt(2) and -sqrt(2) and the eigenvectors are and respectively. You can use the fundamental structure to solve the integrals with matrices if you define the exponental of a square 2x2 matrix using a taylor series type expansion. Ultimately you'd be doing the work that you did solving it the way you did. That's why it's so enjoyable to watch it. Great job!
We see that dy/dx= (x+y+1)/(y-x), which is exact and we get x^2-y^2 +2xy + 2x = constant. Alternatively, d^2x/dt^2 = dy/dt-dx/dt = 2x+1 > x = -1/2 +A e^(sqrt(2)t) + B e^(-sqrt(2)t) and y = x + dx/dt = -1/2 +(sqrt(2)+1)A e^(sqrt(2)t) - (sqrt(2)-1)B e^(-sqrt(2)t) . We have two first order equations and hence the two constants.
Try: let u = x + 1/2, v= y + 1/2 to get the linear system u' = -u + v and v' = u + v. Then integrate using standard techniques.
And you have assigned arbitrary values for a and b ?
What??? Just differenciate the first eq(eith respect to t) and you get(using the second eq) y''=2y+1,and after solving this simple differential eq for y you get x by x=y'-y-1 (the first eq)😊
linear algebra?
If x = x(t) and y = y(t) we have
(1) dt = dy/(x+y+1)
(2) dt = dx/(y-x)
So, integrating
(1) t = ln(x+y+1) + C1
(2) t = -ln(y-x) + C2
where x and y are "constants" when integrating with respect to y and x.
Then ln(x+y+1) + ln(y-x) = C3
ln [(x+y+1)(y-x)] = C3
(x+y+1)(y-x) = constant
That was the incorrect integration. You can write:
dt = dx / (y - x) = dy / (x + y + 1),
but integration will be like this:
dx / (y - x) = dy / (x + y + 1) = d(x + y) / (2 y + 1),
(2 y + 1) dy = (x + y + 1) d(x + y),
after integrating and "decorating":
x² + 2 x y - y² + 2 x = const.
It's wrong because x is not a constant in terms of y and y is not a constant in terms of x
@@appybane8481 Maybe... but he doesn't say anything... He only says that x and y are function of t...
The simple method before me
First we write the equations
dy/dt=x+y+1
dx/dt=y-x
It is enough to divide both sides of the equations
dy/dt/dx/dt=(x+y+1)/(y-x)
Well, friends, take a look at the left side of the equation According to the chain rule, you should be able to simplify it
dy/dt/dx/dt=dy/dx
Now head right First, let's me rewrite the equation once more
dy/dx=(x+y+1)/(y-x)
My dears, is this form of the equation not familiar to you? Can we convert it into a homogeneous equation? How to do this?
Let y=Y+a and x=X+b
so dy=dY. And dx=dX
Let's replace the new variables
dY/dX=(X+Y+a+b+1)/(Y-X+a-b)
We need to homogenize
a+b+1=0 and a-b=0
a=-½ and b=-½
dY/dX=(X+Y)/(Y-X)
We change the variable again. The method of solving the homogeneous equation is known
u=Y+X and v=Y-X
Y=(u+v)/2 and X=(u-v)/2
dY=(du+dv)/2. And dX=(du-dv)/2
We replace the new variables
(du+dv)/(du-dv)=u/v
Using the properties of ratios, we can combine the numerator in the denominator
d(u+v)/(2du)=u/(u+v)
We now have a separable differential equation
(u+v)d(u+v)=2udu
It is not difficult to guess the differentials of each side, but if you like integral, you can get integral
d((u+v)²/2)=d(u²)
d((u+v)²/2-u²)=0
(u+v)²/2-u²=constant
Now we find our initial variables
u=X+Y=x+y+1 v=Y-X=y-x
As results
(2y+1)²/2-(x+y+1)²=const
This differential equation is probably related to a physical problem
Anyway, if we want to find the parameters of the location, i.e. x and y, it is enough to look at the solution of the equation again
(2y+1)²/2-(x+y+1)²=c²
(2y+1)/c*√2))²-((x+y+1)/c)²=1
Similar to this union of pan hyperbolic functions
cosh²(√2t)-sinh²(√2t)=1
Let
(2y+1)/(c*√2))=cosh(√2t)
(x+y+1)/c=sinh(√2t)
y=(c√2cosh(√2t)-1)/2
x=c*sinh(√2t)-c/√2cosh(√2t)-1/2
Don't x and y have to be expressed in terms of t?
@@kevinmadden1645
I have done this too, look at the comment above your comment
is that t same as t in the original equation?
@@appybane8481
Yes, t is the intermediate variable that we call time in physics
Boh...non so se ho capito...y=x+√(2(x^2+x-c))
I obtained the same result by two different methods. I (we) assumed the problem was to find the "path" or "trajectory", y(x).
dy/dx = (x+y+1)/(y-x)
(x+y+1)dx - (y-x)dy = 0
xdx + ydx + dx - ydy + xdy = 0
d(x^2/2) - d(y^2/2) + dx + d(xy) = 0
∴ x^2/2 - y^2/2 + x + xy = c
x^2 - y^2 + 2x + 2xy = c
Can't we divide (dy/dt )/(dx/dt)=(x+y+1)/(y-x)
Yes and this is an exact differential equation, solvable by very standard techniques, to get x^2-y^2 +2xy + 2x = constant.
Is “ y^2 = x^2 + 2xy + 2x + c “ , a simpler answer?