Question

Consider the differential equation x^2 y''-xy'-3y=0. If y1=x3 is one solution use redution of order formula to find a second linearly independent solution

Answer 1

Suppose $y_2(x)=y_1(x)v(x)$ is another solution. Then

$\begin{cases}y_2=vx^3\\{y_2}'=v'x^3+3vx^2//{y_2}''=v''x^3+6v'x^2+6vx\end{cases}$

Substituting these derivatives into the ODE gives

$x^2(v''x^3+6v'x^2+6vx)-x(v'x^3+3vx^2)-3vx^3=0$

$x^5v''+5x^4v'=0$

Let $u(x)=v'(x)$, so that

$\begin{cases}u=v'\\u'=v''\end{cases}$

Then the ODE becomes

$x^5u'+5x^4u=0$

and we can condense the left hand side as a derivative of a product,

$\dfrac{\mathrm d}{\mathrm dx}[x^5u]=0$

Integrate both sides with respect to $x$:

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}[x^5u]\,\mathrm dx=C$

$x^5u=C\implies u=Cx^{-5}$

Solve for $v$:

$v'=Cx^{-5}\implies v=-\dfrac{C_1}4x^{-4}+C_2$

Solve for $y_2$:

$\dfrac{y_2}{x^3}=-\dfrac{C_1}4x^{-4}+C_2\implies y_2=C_2x^3-\dfrac{C_1}{4x}$

So another linearly independent solution is $y_2=\dfrac1x$.