An Interesting System of Differential Equations

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[@onegreengoat9779]

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!

[@user-kp2rd5qv8g]

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.

[@yoav613]

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)😊

[@Jperre2010]

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.

[@tanbw922]

Is “ y^2 = x^2 + 2xy + 2x + c “ , a simpler answer?

[@nalapurraghavendrarao6324]

And you have assigned arbitrary values for a and b ?

[@GeoffryGifari]

linear algebra?

[@FisicTrapella]

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

[@amaxar7775]

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.

[@appybane8481]

It's wrong because x is not a constant in terms of y and y is not a constant in terms of x

[@FisicTrapella]

@@appybane8481 Maybe... but he doesn't say anything... He only says that x and y are function of t...

[@nalapurraghavendrarao6324]

Can't we divide (dy/dt )/(dx/dt)=(x+y+1)/(y-x)

[@user-kp2rd5qv8g]

Yes and this is an exact differential equation, solvable by very standard techniques, to get x^2-y^2 +2xy + 2x = constant.

[@MortezaSabzian-db1sl]

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

[@MortezaSabzian-db1sl]

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

[@kevinmadden1645]

Don't x and y have to be expressed in terms of t?

[@MortezaSabzian-db1sl]

​@@kevinmadden1645 I have done this too, look at the comment above your comment

[@appybane8481]

is that t same as t in the original equation?

[@MortezaSabzian-db1sl]

@@appybane8481 Yes, t is the intermediate variable that we call time in physics

[@user-zp3ch8tt4i]

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

[@giuseppemalaguti435]

Boh...non so se ho capito...y=x+√(2(x^2+x-c))

[@stephenshefsky5201]

I obtained the same result by two different methods. I (we) assumed the problem was to find the "path" or "trajectory", y(x).