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,

`(\mathrm d)/(\mathrm dx)[x^5u]=0`

Integrate both sides with respect to
`x`:

`\displaystyle\int(\mathrm d)/(\mathrm dx)[x^5u]\,\mathrm dx=C`

`x^5u=C\implies u=Cx^(-5)`

Solve for
`v`:

`v'=Cx^(-5)\implies v=-\frac{C_1}4x^(-4)+C_2`

Solve for
`y_2`:

`(y_2)/(x^3)=-\frac{C_1}4x^(-4)+C_2\implies y_2=C_2x^3-(C_1)/(4x)`

So another linearly independent solution is
`y_2=\frac1x`.