Answer to Question #218070 in Differential Equations for Vic

Question

Find the general solution and the second solution of,

x²y"-3xy'+4y=0,given y₁(x)=x²

Answer 1

Answer:-

Given that $y_1(x)=x^2$  is one solution.

$\implies y'_1=2x\\ \implies y''_1=2$

Given D.E "x^2y"-3xy'+4y=0"

Now  $x^2y_1''-3xy'_1+4y_1=x^2(2)-3x(2x)+4x^2=0$

That is. $y_1(x)$ satisfies the given DE

So $y_1(x)$ s a solution of the D.E.

Let the second solution be $y_2(x)=u(x)x^2$

$\implies y_2'=u'x^2+2ux\\\implies y''_2=u''x^2+2u'x+2[u'x+u]$

or $y_2''=u''x^2+4u'x+2u$

Substitute these into the differential equation to get

$x^2(u''x^2+4u'x+2u)-3x(u'x^2+2ux)+4ux^2=0$

$\implies u''x^4+u'x^3=0\\ \implies u''x+u'=0\ \ \ \ \ \ \ \ \ \ \ \ \ (\because x >0) \\ \implies \frac{u''}{u'}=-\frac{1}{x}$

on integrating we get

$lnu'=-lnx+c\\ \ \ \ \ \ \ \ \ =ln(\frac{1}{x})+c$

By taking c=0 we get

$lnu' =ln(\frac{1}{x})\\ \implies u'=\frac{1}{x}\\ \implies u=lnx$

So second solution $y_2(x)=(lnx)x^2$

$W(x)=\begin{vmatrix} x^2 & x^2lnx\\ 2x & 2xlnx+x \end{vmatrix}$

Now wronskian 

$=2x^3lnx+x^3-2x^3$

$=x^3 \\\implies \boxed{W(x)\ne0} \ \ \ \ \ \ (\because x >0)$

Hence $y_1(x) and Y_2(x)$ form a fundamental set of solutions.

So general solution is given by

$\boxed{y=c_1x^2+c_2x^2lnx}$