Asked by Kaneki
Use the theorem of existence and unity to determine if there exists a unique solution for the problem of initial value:
dy/dx= 3y^2/3, y(2)=0
dy/dx= 3y^2/3, y(2)=0
Answers
Answered by
oobleck
y^(-2/3) dy = 3 dx
3y^(1/3) = 3x+c
y = (x-2)^3
Looks good. The Existence Theorem says so too, since
f(x,y) = y - x^3 is continuous at (2,0)
∂f/∂y = 1 is continuous at (2,0)
See this article (section 2.8.1)
www.sciencedirect.com/topics/mathematics/existence-and-uniqueness-theorem
3y^(1/3) = 3x+c
y = (x-2)^3
Looks good. The Existence Theorem says so too, since
f(x,y) = y - x^3 is continuous at (2,0)
∂f/∂y = 1 is continuous at (2,0)
See this article (section 2.8.1)
www.sciencedirect.com/topics/mathematics/existence-and-uniqueness-theorem
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