In Problems 1-10, use a substitution y=xr to find the general solution to the given equation for x>0.
The general solution for the given equation is y=c1 x cos (4 lnx)+c2 x sin(4 lnx)..
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 both examples of Cauchy problems.
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
And let's set
Substituting the w(r,x) in place of y(x), you get
=x2 (r(r-1)) xr-2-x(r)xr-1+17xr
=(r2-r) xr-rxr +17xr
Solving the indicial equation
There are two complex conjugates,
r1=1+4i and r2=1-4i
Thus there are two linearly independent solutions given by
x1+4i=x1 cos(4 ln x)+ix1 sin(4 ln x)
You can write two linearly independent real-valued solutions as,
y1=x cos(4 ln x) and y2=x sin (4 ln x)
The general solution for the equation will be,
y=c1 x cos (4 lnx)+c2 x sin(4 lnx)
Therefore, the general solution for the given equation is y=c1 x cos (4 lnx)+c2 x sin(4 lnx).
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