Question

Find the general solution: (differential eqs - first order)
ty'+2y=4t^2
y(1)=2

Answer 1

i got the answer: y=(t^2)/3 + 5/6

Answer 2

mu(t)= e^integral(2/t)dt

Answer 3

your solution is incorrect.
first divide whole equation with t
y'+2/t y=4t

find integrating factor
\[\Large =e^{\int\limits_{}^{}\frac{2}{t}dt}=e^{2\ln(t)}=e^{\ln(t^2)}=t^2\]
now Multiply whole equation with t^2
\[\Large t^2y'+2ty=4t^3\]
can you do this now ?

Answer 4

wouldnt it be t^2*y=integral(t^2 * 4t * dt)?

Answer 5

nvm ur right

Answer 6

do u mind giving me the answer so i can check at the end

Answer 7

OS coc do not allow to give answers ..
but we can work together to solve this !
let me know where you stuck. m here

Answer 8

ok i integrated the eq u gave me and got
t^3 * y + t^2 * y^2/2= t^4 + C

Answer 9

(ty) + (y''t^2)?

Answer 10

oops!!!
the left side is product rule
\[\Large t^2y'+2ty=4t^3\]
can i write the left side as
\[\Large \frac{d}{dt}(t^2y)=4t^3\]
it it ok ?

Answer 11

yes
if you apply the product rule you will get the same y^2y'+2ty

Answer 12

alright got it thx

Answer 13

you are welcome :)