Question

Pls help:)
Show that \(x^3+3xy^2=1\) gives a solution of the differential equation
\[2xy\frac{dy}{dx}+x^2+y^2=0\] on the interval 0 12 years ago

Answer 1

y^2 = (1-x^3)/(3x)

Answer 2

but thats prolly inconsequential

Answer 3

I have showd that it is a solution.but i dnt knw how to show the interval part.

Answer 4

D[x^3 + 3x y^2 = 1]

D[x^3] + D[3x y^2] = D[1]

3x^2 x' + D[3x y^2] = 0

3x^2 x' + (3x D[y^2] + D[3x] y^2) = 0

3x^2 x' + (3x 2y y' + 3x' y^2) = 0

assumeing its wrt.x; x'=1

3x^2 + (3x 2y y' + 3y^2) = 0

solve for y'

Answer 5

you don't need to solve for y'

Answer 6

2xy not equal to 0

Answer 7

just factor out a 3

Answer 8

but i like solving for y' its my favorite part :)

Answer 9

:) but how do i show that it lies in that interval?

Answer 10

if 2xy not equal to 0, then the rest of it looks fine to me

recall that we would need to divide off that 2xy to get a proper solution for some reason

Answer 11

but i spose that involves finding a y=_____

Answer 12

y^2 = (1-x^3)/(3x)

y =+- sqrt( (1-x^3)/(3x))

so as long as 1-x^3 not equal zero .....

Answer 13

since x = 1 is bad, and x=1 is not in the interval ... its good

Answer 14

i am really sorry. i am new to differential equation. so this interval part is still confusing me.:(

Answer 15

\[ky'+ay=b\]
has a solution (unique solution) if k not equal to 0, why?

Answer 16

\[y'=\frac{b-ay}{k}\]

Answer 17

if k is equal to zero then it becomes undefined which leaves with no solution. is that the reason @amistre

Answer 18

thats a rather simplistic reason yes :) i cant say that im confident for any other reasons ive come across

Answer 19

a solution has to be continuous along an interval, and if the interval has an undefinable point ... then continuity is in question

Answer 20

oh. so i need to find two points where it continuity is in question. in this case when
x=0
y becomes undefined and when x=1 it becomes 0. simply if we get infinity or 0 thn that is the point where continuity ends. am I right?

Answer 21

you would need to find at most, 1 point, where continuity is in question

we need to look at all the possible situations where we get a "bad math" issue.

x^3+ 3x y^2 = 1 , in the original equation there is no problem

\[2xy\frac{dy}{dx}+x^2+y^2=0\]
\[\frac{dy}{dx}+\frac{x^2+y^2}{2xy}=0\]we have an issue when 2xy = 0; so when x=0, or y=0 is problematic

we know that x=0 is not in the interval, so lets determine of y=0 happens ...

x^3+ 3x y^2 = 1 , solve for y, or y^2 even
\[y^2=\frac{1-x^3}{3x}\]
since there is no x=0 in the interval, that leaves us with 1-x^3 = 0 to eliminate

Answer 22

we have 2 solutions for y, which is what i initially drew up
\[y=\sqrt{\frac{c-x^3}{3x}}\]
\[y=-\sqrt{\frac{c-x^3}{3x}}\]
at x=c we therefore have 2 equally valid solutions and cannot determine a unique solution

Answer 23

since we know the particular solution is c=1 .....

Answer 24

so, in the interval:

0< x < 1

we CAN define a unique solution for all values of x with the given solution, is what i get from it all

Answer 25

I get it now. Thanxxx a lot for ur help.:D

Answer 26

youre welcome :)

Answer 27

@amistre sorry to trouble u. if i only show that x^3+3xy^2=1 gives a solution the d.e is it enough? or should I show what happens when x<0 or x>1?

Answer 28

@amistre64

Answer 29

@Zarkon

Answer 30

since they give you the interval: 0<x<1 it is always good practice to test if the interval is valid. The textbooks tend to mention that after they teach this portion of the material that it is assumed that the solutions exist within their proper intervals

Answer 31

i believe one good way to consider this interval of existence debacle is to relate it to the interval of convergence of a power series.