what y is unique in IVP?y'+(1/tlnt)y=9t^2

Expert Answers
sciencesolve eNotes educator| Certified Educator

You need to remember the standard form of IVP such that:

`y' + p(x)y = q(x)`

Comparing the standard form to the equation `y'+(1/(tlnt))y=9t^2`  yields: `p(x)=1/(tlnt); q(x)=9t^2`

You need to consider the integrating factor `I(t)=ln t` , hence you need to multiply the equation by ln t both sides:

`y'*ln t + y/t = 9t^2*ln t`

Notice that `y'*ln t + y/t = (y*ln t)'`

Hence, writing the new form of equation yields:

`(y*ln t)' = 9t^2*ln t`

Integrating both sides yields:

`y*ln t = int 9t^2*ln t dt`

You need to remember the formula of integration by parts to solve `int 9t^2*ln t` `dt`  such that:

`int udv = uv - int vdu`

Considering `u =ln t=gt du = (dt)/t`

Considering `dv = t^2dt =gt v = t^3/3`

Hence, substituting  `ln t, (dt)/t, t^3/3`  for u,du and v yields:

`int 9t^2*ln t dt =9*(t^3*ln t)/3 - (9/3)int t^2 dt`

`int 9t^2*ln t dt =3*(t^3*ln t) - (t^3)+c`

Solving the equation for y yields:

`y = 3*t^3 - (t^3)/(ln t)`

Hence, the unique solution to IVP is `y(t) = 3*t^3 - (t^3)/(ln t).`