Question

Differential Equations Question (First Year Engineering Calculus II)?

Thanks!

Answer 1

`y={( (t^2 )/(2(1 + t^2)) " " 0 le t le 1 ),( ( 2 - t^2)/(2(1 + t^2)) " " t gt 1):}`

Explanation:

`f(t)={(" "t " " 0 le t le 1 ),(-t " " t gt 1):}`

Your LHS is already in exact form:

• `((1 + t^2) y )^' = (1 + t^2) \ y^' + 2t \ y`

For `0 le t le 1`

`((1 + t^2) y )^' = t`

`(1 + t^2) y = t^2/2 + C`

`y(0) = 0 implies C = 0`

`y = (t^2 )/(2(1 + t^2)) qquad square`

For `t ge 1`

`((1 + t^2) y )^' =- t`

`(1 + t^2) y = -t^2/2 + C`

New IV from `square`:

`y(1) = 1/4 implies C = 1`

`y = ( 2 - t^2)/(2(1 + t^2))`

`y={( (t^2 )/(2(1 + t^2)) " " 0 le t le 1 ),( ( 2 - t^2)/(2(1 + t^2)) " " t gt 1):}`