- #1
JamesEllison
- 13
- 0
Problem: Solve the initial Value:
when x=1, y=0
dy/dx = 1
2x^2(d^2y/dx^2) + 3x (dy/dx) - 15y = 0
My attempt:
x = e^t
dx/dt = e ^t
dy/dt = dy/dx * dx/dt
dy/dt = x*dy/dx
d^2y/dt^2
= d/dt(dy/dt)
= d/dt(x*dy/dx)
=d/dx(x*dy/dx)*(dx/dt)
since dx/dt = x
=(x^2*d^2y/dx^2) + (x*dy/dx)
=(x^2*d^2y/dx^2) + (dy/dt)
(x^2*d^2y/dx^2) = (d^2y/dt^2) - (dy/dt)
From this point, do I simply substitute the Values for the 1st/2nd derivatives in ? i.e.
2((d^2y/dt^2) - (dy/dt)) + 3(dy/dt) - 15y = 0
2d^2y/dt^2 - 2dy/dy +3dy/dt - 15y = 0
2m^2 +3m - 1 = 0
Then solve for m,
Then create the solution in the form of :
Ae^mt + Be^mt = 0 ??
Is that the right path? Any help is appreciated.
Cheers.
when x=1, y=0
dy/dx = 1
2x^2(d^2y/dx^2) + 3x (dy/dx) - 15y = 0
My attempt:
x = e^t
dx/dt = e ^t
dy/dt = dy/dx * dx/dt
dy/dt = x*dy/dx
d^2y/dt^2
= d/dt(dy/dt)
= d/dt(x*dy/dx)
=d/dx(x*dy/dx)*(dx/dt)
since dx/dt = x
=(x^2*d^2y/dx^2) + (x*dy/dx)
=(x^2*d^2y/dx^2) + (dy/dt)
(x^2*d^2y/dx^2) = (d^2y/dt^2) - (dy/dt)
From this point, do I simply substitute the Values for the 1st/2nd derivatives in ? i.e.
2((d^2y/dt^2) - (dy/dt)) + 3(dy/dt) - 15y = 0
2d^2y/dt^2 - 2dy/dy +3dy/dt - 15y = 0
2m^2 +3m - 1 = 0
Then solve for m,
Then create the solution in the form of :
Ae^mt + Be^mt = 0 ??
Is that the right path? Any help is appreciated.
Cheers.
Last edited: