Question

Find y as a function of x if
`x^(2) y''-4xy'-6y= x^(3)`
y(1)=6 and y'(1)=-1

Answer 1

Let
`y(x)=z(t)`, where
`t=\ln x`. Then


`\implies y'=\frac1xz'`

`\implies y''=\frac1{x^2}(z''-z')`

so the ODE is equivalent to


`(z''-z')-4z'-6z=e^(3t)`

`z''-5z'-6z=e^(3t)`

The characteristic equation is


`r^2-5r-6=(r-6)(r+1)=0\implies r=6,r=-1`

so that the characteristic solution is


`z_c=C_1e^(6t)+C_2e^(-t)`

For the particular solution, take


`z_p=ae^(3t)`

`\implies {z_p}'=3ae^(3t)`

`\implies {z_p}''=9ae^(3t)`


`\implies 9ae^(3t)-15ae^(3t)-6ae^(3t)=e^(3t)`

`\implies -12a=1`

`\implies a=-\frac1{12}`


`\implies z_p=-\frac1{12}e^(3t)`


`\implies y_p=-\frac1{12}x^3`


`\implies y=y_c+y_p=C_1x^6+\frac{C_2}x-\frac1{12}x^3`

With the initial conditions, we get


`y(1)=6\implies 6=C_1+C_2-\frac1{12}`

`y'(1)=-1\implies -1=6C_1-C_2-\frac14`

`\implies C_1=(16)/(21),C_2=(149)/(28)`

So the particular solution to the IVP is


`y=(16x^6)/(21)+(149)/(28x)-(x^3)/(12)`