Question

Find the solution of the given initial value problem.


\(ty\prime +2y=t^2-t+1\)
\(y(1)=0\)

I have a rough idea of how to go about this, but I am still feeling out how is best to start about solving each type of problem.

Answer 1

Ok, so i need to get it in the standard for yeah?

Answer 2

\(y\prime+\frac{2}{t}y=t-1+\frac{1}{t}\)
That look ok?

Answer 3

multiply both sides by t
and you will get \[\frac{ d }{ dt }\left( t ^{2} y\right)\] on the left side

Answer 4

and then seperation of variables

Answer 5

first solve hom. part\[\Large ty'+2y=0

\]this is separable..

Answer 6

for particular sol. search by assuming y(t)= At^2+Bt+C is sol..

Answer 7

\[\Large ty'+2y=0\\


\Large ty'=-2y\\

\Large\frac{dy}{y}=\frac{-2}{t}dt
\]

Answer 8

if you multiply both sides by t \[t ^{2}\frac{ dy }{ dt }+2ty=t ^{3}-t ^{2}+t\]\[t ^{2}\frac{ dy }{ dt }+\frac{ d \left( t ^{2} \right) }{ dt }y=t ^{3}-t ^{2}+t\]\[\frac{ d \left( t ^{2}y \right) }{ dt }=t ^{3}-t ^{2}+t\]\[d \left( t ^{2}y \right)=\left( t ^{3}-t ^{2}+t \right)dt\]

Answer 9

now integrate both sides and you will get y(t)