Questions - Differential Equations

Questions / Answers
Differential Equations

On 10/23/06, Anonymous wrote:
> Anonymous has left a new comment on your post "Questions - Differential
> Equations":
>
> Can someone help me with this? For a second order differential equation of
> the form y" = f(x,y'), the substitutions v = y', v' = y" lead to a first
> order differential equations of the form v' = f(x,v). Provided that this is
> differential equation for v can be solved, y can be obtained by integrating
> dy/dx = v(x) Notice that one arbitary constant is introduced in solving the
> differential equation for v and that another is introduced when integrating
> v(x) to obtain y. Solve the following differential equation x^2y" + 2xy' - 1
> = 0
>
>
>
> Posted by Anonymous to Maths,management,counselling at 9:26 AM
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x^2y"+2xy'-1=0
put
x=e^z....dx/dz=e^z=x.....dz/dx=1/x
y' = dy/dx = (dy/dz)(dz/dx)=Dy/x..where Dy represents dy/dz
y"= d^2y/dx^2 = (d/dx){dy/dx}=(d/dx){Dy/x}
=(1/x^2)[(d^2y/dz^2)-(dy/dz)]=(1/x^2)[D^2y-Dy]
=(1/x^2)[D(D-1)]y
so the given eqn. becomes
D(D-1)y+2Dy-1=0
[D^2-D+2D]y = 1
[D^2+D]y=1
D(D+1)y=1
c.f
y=c1e^(0z)+c2e^(-1z)......
p.i........
[1/D(D+1)]1...etc
this is a standard differential eqn. in z which you can solve for y.then substitute e^z =x
or z=ln(x)