(dy)/(dt) + 0.5yt=8t and y(0)=9 Find y(t)

Question

`(dy)/(dt) + 0.5yt=8t` and
`y(0)=9`

Find y(t)

Answer 1

`(\mathrm dy)/(\mathrm dt)+\frac t2y=8t`

`e^(t^2/2)(\mathrm dy)/(\mathrm dt)+\frac t2e^(t^2/2)y=8te^(t^2/2)`

`(\mathrm d)/(\mathrm dt)\left[e^(t^2/2)y\right]=8te^(t^2/2)`

`e^(t^2/2)y=8\displaystyle\int te^(t^2/2)\,\mathrm dt`

`e^(t^2/2)y=8e^(t^2/2)+C`

`y=8+Ce^(-t^2/2)`

With
`y(0)=9`, we get


`9=8+Ce^0\implies C=1`

and so the particular solution must be


`y=8+e^(-t^2/2)`