Solving the IVP (x - 3) y1 + y = sqrt[x] (set-up help)

[forza1nter]

k, here is a prob my prof gave in class and the most annoying thing is setting up the prob to intergrate....

this is driving me nuts b/c i know its just a matter of algebra and its just not clicking...thanks in advance!!!

my work thus far:

(x - 3)y' + y = sqrt[x]
(x - 3)dy/dx + y = sqrt[x]
(x - 3)dy + y = sqrt[x]dx
dy + y = sqrt[x]dx/(x - 3) - integrate

the LHS (y's) integrate nice, but the RHS (x's) I get a weird, long process...

 

[royhaas]

Re: Solving the IVP....setup help!!

Start by checking your algebra.

 

[Deleted member 4993]

Re: Solving the IVP....setup help!!

k, here is a prob my prof gave in class and the most annoying thing is setting up the prob to intergrate....

this is driving me nuts b/c i know its just a matter of algebra and its just not clicking...thanks in advance!!!

my work thus far:

(x - 3)y' + y = sqrt[x]
(x - 3)dy/dx + y = sqrt[x]

Convert the above to:

\(\displaystyle y' \, + y \cdot \frac{1}{x-3} = \frac{\sqrt{x}}{x-3}\)

This is of the form

\(\displaystyle y' + y\cdot p(x) \, = \, g(x)\)

To solve these you need multiply both sides by

\(\displaystyle e^{\int{p} \, dx}\)

and continue....

(x - 3)dy + y = sqrt[x]dx
dy + y = sqrt[x]dx/(x - 3) - integrate

the LHS (y's) integrate nice, but the RHS (x's) I get a weird, long process...