Answer to Question #215068 in Differential Equations for Prince

Question

Given that y1=√x lnx is a solution of
4x^2 y"+y=0
Use the reduction of order to obtain the general solution on (0, ∞) show that tge set of solutions (y1, y2) is a fundamental set

Answer 1

Let $y=y_1u$ , hence $y'=y_1'u+y_1u'$ , and $y''= y_1''u+2y_1'u'+y_1u''$.

Then

$0=4x^2 y''+y=4x^2 (y_1''u+2y_1'u'+y_1u'') +y_1u =$

$= 4x^2(2y_1'u'+y_1u'')+(4x^2y_1''+y_1)u=4x^2(2y_1'u'+y_1u'')$, since $4x^2y_1''+y_1=0$.

$2y_1'u'+y_1u''=0$

$2y_1'y_1u'+y_1^2u''=(y_1^2u')'=0$

$y_1^2u'=C_1$

$u'=C_1/y_1^2$

$u=\int\frac{C_1dx}{x\ln^2 x} =C_0 - C_1\frac{1}{\ln x}$

$y=y_1u=(C_0 - C_1\frac{1}{\ln x})\sqrt{x}\ln x=C_0\sqrt{x}\ln x-C_1\sqrt{x}$ is the general solution.

The functions $y_1(x)=\sqrt{x}\ln x$, $y_2(x)=\sqrt{x}$ are linearly independent. Indeed, if

$a_1y_1+a_2y_2= 0$ , then $a_1\ln x+a_2=0$ . Since $a_1\ln x=-a_2$ is a constant, $a_1=0$ and hence $a_2=0$. This means that $y_1(x)$ and $y_2(x)$ are linearly independent.

The functions $y_1(x)$ and $y_2(x)$ form a fundamental set of solutions, since they are linear independent, and the general solution is their arbitrary linear combination.