In Problems 15-17,solve the given initial value problem x2y"+5xy'+4y=0.
y(1) =3 and y'(1) = 7
The solution of the given initial value problem is y=3x-2+13x-2 ln x.
In mathematics, a Cauchy problem is one in which the solution to a partial differential equation must satisfy specific constraints specified on a hypersurface in the domain.
An initial value problem or a boundary value problem is an example of the Cauchy problem.
The equation will be in the form of, ax2y"+bxy'+cy=0.
The given equation is,
Let L be the differential operator defined by the left-hand side of equation, that is,
L [x] (t) = x2y"+5xy'+4y
w (r,t) = xr
Substituting the w(r,t) in place of x(t), you get
L [w] (t)=x2 (xr)"+5x (xr)'+4 (xr)
=x2 (r (r-1)) xr-2 +5x (r) xr-1 +4xr
= (r2-r) xr +5rxr +4xr
Solving the indicial equation,
There are repeated roots at r= -2
Thus there are two linearly independent solutions given by
y1=c1x-2 and y2 = c2x-2 lnx
The general solution for the equation will be
y=c1x-2 +c2x-2 lnx
For the given initial conditions,
y(1)=3 and y'(1)=7
y(x)=c1 x-2+c2 x-2 lnx
y(1) = c1
y'(x)=c1 (-2) x-3 +c2 [(-2) x-3 lnx+x-2 (1/x)]
y'(1) = -2c1+c2
Solving the two simultaneous equations (1) and (2), you get the values of two constants c1 and c2 as,
c1 = 3 and c2 = 13
Thus the solution of the given initial value problem is y=13 x-2+13x-2 ln x.
where n is an unspecified parameter is called Legendre’s equation. This equation appears in applications of differential equations to engineering systems in spherical coordinates.
(a) Find a power series expansion about x=0 for a solution to Legendre’s equation.
(b) Show that for a non negative integer there exists an nth degree polynomial that is a solution to Legendre’s equation. These polynomials upto a constant multiples are called Legendre polynomials.
(c) Determine the first three Legendre polynomials (upto a constant multiple).
For Duffing's equation given in Problem 13, the behaviour of the solutions changes as r changes sign. When , the restoring force becomes stronger than for the linear spring. Such a spring is called hard. When , the restoring force becomes weaker than the linear spring and the spring is called soft. Pendulums act like soft springs.
(a) Redo Problem 13 with . Notice that for the initial conditions , the soft and hard springs appear to respond in the same way for small.
(b) Keeping and , change the initial conditions to and . Now redo Problem 13 with .
(c) Based on the results of part (b), is there a difference between the behavior of soft and hard springs for small? Describe.
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