Question

Find the general solution of the differential equation y^(5) −2y^(4) + y^(3) = 0.

Answer 1

`y^((5))-2y^((4))+y^((3))=0`

We can reduce the order of the ODE by substituting
`v(x)=y^((3))(x)`, so that
`v'(x)=y^((4))(x)` and
`v''(x)=y^((5))(x)`. Then

`v''-2v'+v=0`

has characteristic equation

`r^2-2r+1=(r-1)^2=0`

with root
`r=1`, which has multiplicity 2, so that the characteristic solution is

`v_c=C_1e^x+C_2xe^x`

Integrate both sides to solve for
`y''(x)`:

`y''=C_1e^x+C_2e^x(x-1)+C_3`

`y''=C_1e^x+C_2xe^x+C_3`

Integrate again to solve for
`y'(x)`:

`y'=C_1e^x+C_2e^x(x-1)+C_3x+C_4`

`y'=C_1e^x+C_2xe^x+C_3x+C_4`

And one last time to solve for
`y(x)`:

`y=C_1e^x+C_2e^x(x-1)+\frac{C_3}2x^2+C_4x+C_5`

`\boxed{y(x)=C_1e^x+C_2e^x+C_3x^2+C_4x+C_5}`