Answer to Question #160351 in Differential Equations for Saif

Question

Solve the following equations using Wronskian’s method

x²d²y/dx²-3xdy/dx+40y=x²sin(logx)

Answer 1

Consider Euler equation:

$x^2y''-3xy'+40y=0$

This is Euler Equation.

Let $y(x)=x^r$ ; then:

$r(r-1)-3r+40=0$

$r^2-4r+40=0$

$r_{1,2}=\frac{4-\sqrt{16-160}}{2}=2\pm6i$

The general solution is

$y(x)=y_1(x)+y_2(x)=x^2(c_1cos(6lnx)+c_2sin(6lnx))$

Now consider equation

$y''-\frac{3}{x}y+\frac{40}{x^2}=sin(lnx)$

This is equation of the form:

$y''+a(x)y'+b(x)y=f(x)$

The Wronskian is

$W(x)=c\cdot exp\{-\intop a(x)dx\}$

$W(x)=ce^{-3lnx}=c_3x$

The particular solution:

$y_p(x)=-y_1(x)\intop \frac{y_2(x)f(x)}{W(x)}dx+y_2(x)\intop \frac{y_1(x)f(x)}{W(x)}dx$

$\intop \frac{y_2(x)f(x)}{W(x)}dx=\frac{c_2}{c_3}\intop\frac{x^2sin(6lnx)sin(lnx)}{x}dx=$

$=\frac{c_2}{c_3}\frac{x^2(265sin(5lnx)-203sin(7lnx)+106cos(5lnx)-58cos(7lnx))}{3074}$

$\intop \frac{y_1(x)f(x)}{W(x)}dx=\frac{c_1}{c_3}\intop\frac{x^2cos(6lnx)sin(lnx)}{x}dx=$

$=\frac{c_1}{c_3}\frac{x^2(-106sin(5lnx)+58sin(7lnx)+265cos(5lnx)-203cos(7lnx))}{3074}$

The general solution of the initial equation:

$y(x)=y_1(x)+y_2(x)+y_p(x)$