Question

x²(d²y/dx²) + x(dy/dx) - 4y = x³

Answer 1

Given:

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

This is an Euler-Cauchy equation with variable coefficients.

Let's try to solve the following homogeneous equation without y-free part (right hand side):

$x^2y''+xy'-4y=\bold{0}$

Try to find any trivial solution in form:

$y=x^a, a - const$

$y'=ax^{a-1}, y'' = a(a-1)x^{a-2}$

$x^2(a(a-1)x^{a-2})+x^2(ax^{a-1})-4x^a=0$

$a(a-1)x^a+ax^a-4x^a=0$

$a^2-a+a-4=0,\space a =\pm2$

We found two trivial solutions, let's choose the first one:

$1. \space y=x^2,\space 2.\space y=x^{-2}$

Now try to find a general solution in form:

$y=x^2u(x)=x^2u$

$y'=2xu+x^2u'$

$y''=2u+4xu'+x^2u''$

$x^2(\blue{2u}+4xu'+x^2u'') + x(\blue{2xu}+x^2u')-\blue{4x^2u}=x^3$

$\blue{4x^3u'}+\underline{x^4}u''+\blue{x^3u'}=x^3$

$5\blue{x^3}u'+\underline{\blue{x^3}x}u''=\blue{x^3}$

$5u'+xu''=1\space(*)$

The equation (*) is the first-order linear differential equation.

Let's solve first homogeneous variant of this equation:

$5u'+xu''=\bold{0}$

To simplify the solution we can use substitution:

$u'(x)=t(x)=t, \space t'=\frac{dt}{dx}=u''$

$5t+xt'=0$

$x\frac{dt}{dx}=-5t$

$\frac{dt}{t}=-5\frac{dx}{x} \space (integrate \space both \space parts)$

$ln|t|=-5ln|x|+ln|C|, \space C - const$

$t=\frac{C}{x^5}$

Now we can use the method of variation of parameters to find the general solution of (*):

$t=\frac{C(x)}{x^5}, \space t'=\frac{C'(x)}{x^5}-5\frac{C(x)}{x^6}$

$\blue{5\frac{C(x)}{x^5}}+x\frac{C'(x)}{x^5}-\blue{5x\frac{C(x)}{x^6}}=1$

$x\frac{C'(x)}{x^5}=1$

$C'(x)=x^4 \space (integrate \space both \space parts)$

$C(x)=\frac{x^5}{5}+D, \space D - const$

$t=\frac{C(x)}{x^5}=\frac{x^5+D}{x^5}=\frac{1}{5}+\frac{D}{x^5}$

$t=u'=\frac{1}{5}+\frac{D}{x^5} \space (integrate \space both \space parts)$

$u=\frac{x}{5}-\frac{D}{4x^4}+E, \space E - const$

$y=x^2u=x^2(\frac{x}{5}-\frac{D}{4x^4}+E)=\frac{x^3}{5}-\frac{D}{4x^2}+Ex^2$

Now we can rewrite constants:

$C_1=-\frac{D}{4}, \space C_2=E$

The general solution of the given equation:

$y=\frac{x^3}{5}+\frac{C_1}{x^2}+C_2x^2, \space C_1,C_2 - const$