Question

Show that the function y(t)= 1/3t^2 +c/t is a solution to the different eqution ty' +y=t^2, then find the particular solution that corresponds to the intial condition y(1)=2

Answer 1

@Hero

Answer 2

you can plug the solution into the DE

so calculate \(t(\frac{1}{3}t^2 +\frac{c}{t})' + (\frac{1}{3}t^2 +\frac{c}{t})\)

, or you can solve this directly using, say, an integrating factor.

Answer 3

sorry I still do not understand it.

Answer 4

@IrishBoy123

Answer 5

starting with \(y = \frac{1}{3}t^2 +\frac{c}{t}\)
calculate \(y' = ...\)

then evaluate ty' + y

[alternatively note that \((ty)' = ty' + y\) ... but i suspect you should ignore that thought and just go as advised]

Answer 6

oay so y'= 2/3t^2 -c/t^2

Answer 7

what will be my ty' in this case?? @IrishBoy123

Answer 8

well your y' is just a little off

Answer 9

\[y=\frac{1}{3}t^{2}+\frac{c}{t}=\frac{1}{3}t^2+ct^{-1} \\ y'=\frac{1}{3} \cdot (2t)+c \cdot (-1 t^{-2}) =\frac{2}{3}t-\frac{c}{t^2}\]

Answer 10

anyways to find ty' you just multiply t to the y' there

Answer 11

\[t y'=t(\frac{2}{3}t-\frac{c}{t^2}) \text{ you can distribute }\]

Answer 12

and then of course you add that that to your y you have
and see if the result is t^2

Answer 13

ty'= 2/3 t^2 - c/t^3 + y
where Y is 1/3t^2 +c/t
4/3t^2 -c/t^3+c/t

Answer 14

t/t^2 is 1/t
not 1/t^3

Answer 15

oh i see so it would be +c/t -c/t so that term is cancled

Answer 16

and am left with 4/3t^2

Answer 17

\[y=\frac{1}{3}t^2+\frac{c}{t} \\ y'=\frac{2}{3}t-\frac{c}{t^2} \\ ty'=\frac{2}{3}t^2-\frac{c}{t} \\ \text{ so } ty'+y=\frac{2}{3}t^2-\frac{c}{t}+\frac{1}{3}t^2+\frac{c}{t}\]

Answer 18

well 2+1 is 3 not 4

Answer 19

so it should be 3/3 t^2

Answer 20

sorry. okkay so I am left with 3/3t^2 which is t^2

Answer 21

right which means that thing that they called y is like totally a solution to the differential equation given

Answer 22

since both sides of the diff equation are equal when we plug that in

Answer 23

right but at the end the questin has another part which is then find the particular soulution.... y(1)=2

Answer 24

then you can find c by using the condition given

Answer 25

so I plug in 1 where t is in the original equation?

Answer 26

yep and after that replace y(1) since you replace all the t's with 1
...replace y(1) with 2
and solve equation for c

Answer 27

\[y(1)=\frac{1}{3}(1)^2+\frac{c}{1}\]
replaced t's with 1
now replace y(1) with 2 since y(1)=2

Answer 28

2=1/3 +c
2-1/3=c
6-1/3=c
5/3=c

Answer 29

that's right
so the particular solution they are looking for is
y(t)=1/3 t^2+5/(3t)

Answer 30

I just plugged that value for c back into y(t)=1/3t^2+c/t

Answer 31

right so my final answer would be y(t)=1/3 t^2+5/(3t)

Answer 32

yep for the particular solution part of the question

Answer 33

alright yaya done with this! tysm u are really good and thaks for taking the tiem to type u everyting in equation format I love it when ppl at openstudy teach me with that. TYSM :)

Answer 34

well some of it I was lazy on and didn't type in the thingy

Answer 35

and that fac tthat It takes a while to type up everything in equation format. NO, u did good for most of the part

Answer 36

k thanks
glad you understand the problem though :)

Answer 37

I understand it if it's less sentence and more math. but thank you for helping me!