Question

\[\frac{dy}{dx} + \frac{3y}{x} + 2x = x^2 + 4 \quad ; \quad y(1) = 0\]

i got \[y = \frac{x^3}6 - \frac{2x^2}5 + x - \frac{23}{30}\]
is that right?

Answer 1

y(1)=0 not satisfied.

Answer 2

why not?

Answer 3

put x=1,do u get y=0 ?

Answer 4

hmm mus've done something wrong

Answer 5

oh,it does.,constant is correct.

Answer 6

i am not satisfied either

Answer 7

oh wait...seems my first answer was right

Answer 8

what about the solution? is it right?

Answer 9

http://www.wolframalpha.com/input/?i=dy%2Fdx%2B3y%2Fx%2B2x%3Dx%5E2%2B4+for+y

Answer 10

i have no idea what the heck that thing means

Answer 11

only term incorrect=c1/x^3
it'll be 23/(30x^3)

Answer 12

i see. thanks