Question

Find solutions to the equation:
xy′+2y=x^3
such that y = 1 when x = 1

Answer 1

first divide through by x
y' + 2 y/x = x^2
you now have the equation in a form where you can apply an integrating factor

Answer 2

so if I say y = uv and y' = u'v+uv' and put that in I get:
u'v+uv' + (2uv)/x = x^2

Answer 3

Is that what you mean by integrating factor?

Answer 4

need any more help?

Answer 5

Yes please

Answer 6

ok integrating factor = e ^ INT 2/x dx = e ^ ln x^2 = x^2

Answer 7

so y IF = INT Q.IF dx where IF = integrating factor and Q = x^2

Answer 8

x^2 *y = INT x^2 * x^2 dx

Answer 9

can you carry on from here?

Answer 10

ok y(x^2) = (x^5)/5 + c

Answer 11

right!

Answer 12

now divide by x^2

Answer 13

y=x^3/5 + c/(x^2)

Answer 14

good job

Answer 15

then plug the values in I get 1 = (1^3)/5 + c/(1^2)
1 = 1/5 + c
c = 4/5

Answer 16

y=(x^3)/5+4/(5x^2)

Answer 17

right

the two things to remember with these equations where you cant separate the variables is
rearrangge them to the form
y' + Py = Q

and use y*IF = INT Q*IF dx

you can look at it as reverse of Product Rule if u like but for me the easiest way is as I've mentioned

Answer 18

Thanks a lot for your help

Answer 19

no probs