Question

X ^ (2) y '' - 7xy '+ 16y = 0, y1 = x ^ 4

Answer 1

Standard reduction of order procedure: suppose there is a second solution of the form
`y_2(x)=v(x)y_1(x)`, which has derivatives


`y_2=vx^4`

`{y_2}'=v'x^4+4vx^3`

`{y_2}''=v''x^4+8v'x^3+12vx^2`

Substitute these terms into the ODE:


`x^2(v''x^4+8v'x^3+12vx^2)-7x(v'x^4+4vx^3)+16vx^4=0`

`v''x^6+8v'x^5+12vx^4-7v'x^5-28vx^4+16vx^4=0`

`v''x^6+v'x^5=0`

and replacing
`v'=w`, we have an ODE linear in
`w`:


`w'x^6+wx^5=0`

Divide both sides by
`x^5`, giving


`w'x+w=0`

and noting that the left hand side is a derivative of a product, namely


`(\mathrm d)/(\mathrm dx)[wx]=0`

we can then integrate both sides to obtain


`wx=C_1`

`w=\frac{C_1}x`

Solve for
`v`:


`v'=\frac{C_1}x`

`v=C_1\ln|x|+C_2`

Now


`y=C_1x^4\ln|x|+C_2x^4`

where the second term is already accounted for by
`y_1`, which means
`y_2=x^4\ln x`, and the above is the general solution for the ODE.