Question

Find the initial value problem of: dy/dx - 2xy = 2xe^(x^2)

Answer 1

are u familiar with linear differential equations of the form
\[\frac{ dy }{ dx }+Py=Q\]

Answer 2

we can solve if by finding IF(integrating factor)

Answer 3

I tried to solve it that way but i'm stuck.. Can you show me how to solve this problem?

Answer 4

yeah
\[\Large \frac{ dy }{ dx }+(-2x)y=2xe^{x^2}\]
i can write it in this way okay?

Answer 5

yes

Answer 6

so now comapring with the above equation we can write
\[P=-2x\]and
\[Q=2xe^{x^2}\]
right!!

Answer 7

still correct ;)

Answer 8

now are u familiar with this?\[I.F=e^{\int\limits_{}^{}Pdx}\]

Answer 9

yes i have heard of it but never applied it

Answer 10

okay so here we need to apply this method using integrating factor
so
\[\Large I.F=e^{\int\limits_{}^{}Pdx}=e^{\int\limits_{}^{}-2xdx}=e^{-x^2}\]
okay?

Answer 11

yes! thats what i found

Answer 12

al right
now consider this equation again\[\frac{ dy }{ dx }+Py=Q\]
now after finding I.F ,to solve the diff.equation u have to do this----
\[y*(I.F)=\int\limits_{}^{}Q*(I.F)dx+C\]

Answer 13

so we have the \[I.F=e^{-x^2}\]
right?
now just apply it

Answer 14

\[y*(e^{-x^2}=\int\limits_{}^{}2xe^{x^2}*(e^{-x^2})dx+C\]
ok getting this?

Answer 15

yes thats right

Answer 16

sorry missed a braket in the lefthand side

Answer 17

so now it will be simply
\[y*e^{-x^2}=\int\limits_{}^{}2xdx+c\]
now i think u can solve it

Answer 18

cAn u?

Answer 19

yes i can! Great. Thank for your help!!!

Answer 20

no problem buddy
and u can click the "best response" button if u're satisfied with my answer :)
have a nice day