# P3 Differential EquationsWatch

This discussion is closed.
#1
Does solving differential equations mean finding y say in the following example?

dy/dx=x

And how do you solve this one

dy/dx=x^+y^2?
0
14 years ago
#2
Yes, solving the DE means finding y=

And the second if its meant to read dy/dx = x^2 + y^2, its not analytically soluable (according to my DE text book!)
0
#3
(Original post by [email protected])
Yes, solving the DE means finding y=

And the second if its meant to read dy/dx = x^2 + y^2, its not analytically soluable (according to my DE text book!)
Thanks

Are any of them hard?
0
14 years ago
#4
isn't the second one simply -y^-1= 1/3 x^3? Split the variables intergrate both sides and voila
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#5
(Original post by BloodyValentine)
isn't the second one simply -y^-1= 1/3 x^3? Split the variables intergrate both sides and voila
No....I dont think so.....have you forgotten to divide x^2 by y^2?
0
14 years ago
#6
the second one wasnt fully written out by the author, could be solvable but from what it looks like its a first order differential equation where u need an intergrating factor, P4 baby
0
14 years ago
#7
dy/dx = x^2 + y^2

The solutions to this ODE are givin as Bessel functions, namely BesselJ and BesselY:

y = -x*(c*BesselJ(-3/4, (x^2)/2) + BesselY(-3/4, (x^2)/2))/(c*BesselJ(1/4, (x^2)/2) + BesselY(1/4, (x^2)/2))

Where c is an arbitrary constant.

BesselJ and BesselY are the bessel functions of the first and second kinds, respectively. They satisfy Bessel's equation:

(x^2)*d/dx[dy/dx] + x*[dy/dx] + (x^2 - v^2)y = 0

Euclid
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