x2y′′−3xy′+4y=x3

The given equation is a Cauchy-Euler differential equation. The general form of a Cauchy-Euler equation is:

ax^2y'' + bxy' + c*y = f(x)

where a, b, and c are constants, y'' is the second derivative of y with respect to x, y' is the first derivative of y with respect to x, and f(x) is a function of x.

The given equation is:

x^2y'' - 3xy' + 4y = x^3

Comparing this with the general form, we have a=1, b=-3, c=4, and f(x)=x^3.

The solution to a Cauchy-Euler equation is of the form:

y = x^r

where r is a root of the characteristic equation:

ar(r-1) + b*r + c = 0

Substituting a=1, b=-3, and c=4 into the characteristic equation gives:

r*(r-1) - 3*r + 4 = 0

Solving this quadratic equation for r gives the roots r1 and r2. These roots give the complementary solution to the homogeneous equation (the equation set equal to zero):

y_c = C1x^r1 + C2x^r2

where C1 and C2 are constants.

The particular solution to the nonhomogeneous equation (the original equation) can be found using a method such as undetermined coefficients or variation of parameters.

The general solution to the differential equation is the sum of the complementary solution and the particular solution.