Question:
Consider xdy = 3ydx.
(a) Apply Theorem 1 and show the equation has unique solution in xy-plane where x ≠ 0
(b) Theorem 1 will be inconclusive about existence and uniqueness of solution where x = 0 (that is on y-axis). For points on the y-axis (i) Show there are infinitely many different solutions for xdy = 3ydx y(0) = 0. (ii) Show there is no solution for xdy = 3ydx,
y(0) = b, b ≠0.
Theorem 1 is attached.
Answer:
Given:
xdy = 3ydx
a) Given theorem 1 :
dy/dx = f (x, y)
we have:
[tex] f(x, y) = \frac{3y}{x}[/tex]
[tex] \frac{d}{dy} f(x, y) = \frac{3}{x}[/tex]
[tex] \frac{d}{dy} [/tex] is at interior of all angles except at point x=0.
Therefore for x≠0
[tex] \frac{d}{dy} f(x, y) = \frac{3}{2}[/tex]
Thus, from theorem 1, there is a unique solution in the xy plane where x≠0
b) i) xdy = 3ydx y(0) = 0
We have:
[tex] \frac{dy}{y} = \frac{3}{x} dx [/tex]
Integrating, we have:
∫[tex] \frac{dy}{y}[/tex] =∫[tex] \frac{3}{x} dx [/tex]
ln y = 3lnx + lnC
y = Cx³
Hence, y(0)=0 is satisfied for all values of C
Thus, there are infinitely many solutions for xdy = 3ydx
ii) xdy = 3ydx
y(0) = b, b ≠ 0.
From our part (i) above, we already know that, y = Cx³
Let's now take initial value of y as b,
ie, y(0) = b
b = 0
From the question, b≠0.
Thus, it means it has no solution.
