MHB What is the integrating factor for solving this IVP?

[karush]

find the solution of the IVP
$ty'+2y=t^2-t+1, \quad y(1)=\dfrac{1}{2}, \quad t>0$\begin{array}{lll}
\textsf{Divide thru by t} & y'+\dfrac{2}{t}y=t-1+\dfrac{1}{t}\\
\textsf{Find u(t)} &u(t)=\exp\displaystyle\int \dfrac{2}{t} \, dx =e^{2ln |t|}+c
\end{array}

so far anyway...
is there some symbol for integrating factor the book seems to use u(t)

 

[skeeter]

Greek letter mu, $\mu$, is what I recall.

for #15, $\mu = t^2$

 

[karush]

Greek letter mu, $\mu$, is what I recall.

for #15, $\mu = t^2$

got it...

 

[HOI]

You have a rather silly algebra mistake in the first step.
$y'+ 2y= te^{-2t}$
Dividing by t gives $\frac{y'+ 2y}{t}= \frac{y'}{t}+ \frac{2y}{t}= e^{-2t}$
not $y'+ \frac{2y}{t}= e^{-2t}$.

Personally, I wouldn't do that. As I said in my response to a previous question, an "integrating factor" is a function, $\mu(t)$, such that $(\mu(t)y)'= \mu(t)y'+ 2\mu(t)y$. Since $(\mu(t)y)'= \mu(t)y'+ \mu'(t)y= \mu(t)y'+ 2\mu(t)y$ so that we must have $\mu'= 2\mu$ so $\mu(t)= e^{2t}$. Multiplying this equation by $e^{2t}$ we get $e^{2t}y'+ 2e^{2t}y= (e^{2t}y)'= t$. Integrating both sides, $e^{2t}y= \frac{1}{2}t^2+ C$ so $y= Ce^{-2t}+ \frac{1}{2}t^2$.

This happens to be a linear differential equation so we can do the "associated homogeneous equation", $y'+ 2y=0$, then the entire equation. The general solution to that homogeneous equation is, of course, $y= Ce^{-2t}$. For the entire equation, with right side, $te^{-2t}, we might try a "particular solution" of the form $Ate^{-2t}$ but since $e^{-2t}$ is already a solution to the homogeneous equation we try, instead, $At^2e^{-2t}$. With $y= At^2e^{-2t}$, $y'= 2Ate^{-2t}- 2Ae^{2t}$ so $y'+ 2y= 2Ate^{-2t}- 2Ae^{-2t}+ 2ate^{-2t}= 2Ate^{-2t}= te^{-2t}$ so $A= \frac{1}{2}$.

 

[karush]

that is #14

 

[Prove It]

View attachment 11295
image to avoid typos... doing #15

\begin{array}{lll}
\textsf{Divide thru by t} & y'+\dfrac{2}{t}y=t-1+\dfrac{1}{t}\\
\textsf{Find u(t)} &u(t)=\exp\displaystyle\int \dfrac{2}{t} \, dx =e^{2ln |t|}+c
\end{array}

so far anyway...
is there some symbol for integrating factor the book seems to use u(t)

It's actually $\displaystyle \begin{align*} \textrm{exp}\left( 2\ln{ \left| t \right| } + C \right) \end{align*}$. Any constant will be fine, so C = 0 is the easiest. It also simplifies to $\displaystyle \begin{align*} \textrm{exp}\left( 2\ln{ \left| t \right| } \right) &= \textrm{exp} \left[ \ln{ \left( \left| t \right| ^2 \right) } \right] = t^2 \end{align*}$.