A Radical Differential Equation | Can You Solve?
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- เผยแพร่เมื่อ 24 ส.ค. 2024
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Why bro. This is perfectly separable
😊
dy/sqrt(y-1)=dx/sqrt(x-1)
sqrt(y-1)=sqrt(x-1)+c
y=1+{x-1 +2c*sqrt(x-1) +c^2}
y=x+2c*sqrt(x-1) +c^2
Separable differential equation
Divide and integrate
This video finds the correct solution for x > 1. The solution for x < 1 is
y = x - 2C(1-x)^(1/2) + C^2, where C is any real.
Genial
Try to solve another DE but this time with the conjugate
dy/dx=sqrt((x-1)/(y-1))
I know it's much easier but it would be fun as a comparison
Good idea!
I tried it out, it is more complicated than it seems
You might consider noting that dy/dx is not a fraction and that the cross multiplication separation only works because of the dx implicit in the applied integral. Treat dy/dx as y' and then separate and then apply the integral.
dy/dx is a fraction, a ratio of two infinitesimals! 😜
With some manipulations i found a relation that imply c1=0 and finally y(x)=x. I don't know if i made a mistake.
I solved for y without expanding, so I could just express c/2 as another constant k.
I cheated and did two substitutions: t = x - 1, and u = y - 1. Then dt = dx, du = dy, and dy/dx = du/dt = sqrt(u/t). Separable, easy to integrate, and then write the result in y and x.
dy / √( y -1) = dx / √ ( x - 1)
√ ( y - 1) = √ ( x - 1) + √ ( y_o - 1)
Here in y_o refers to y at x = 0
Good video 👍🏽
Thank you! 😃
sqrt(a÷b) = sqrt(a)÷sqrt(b) only if sgn(a) = sgn(b)
√(a/b) is equal to √a/√b if and only if a ≥ 0 and b > 0, you should say. If sgn(a) = sgn (b) but both negatives, the √a/√b is not well defined among real numbers; even considering complex numbers, the square root √ becomes a multivalued function.
Like
I don't understand the point of the first solution, it's such an obvious separable equation
just an attempt
Ok, the Solution was so Easy 😅 I figured something More complicated (define new variables: z = y-1; u=x-1).
Excellent!
This is probably the easiest DE you've solved in this channel.
Is it?