Can anyone help to integrate this ? thanks
\Rightarrow \frac {dy}{(y + 3)^2} = (x - 1)^2~dx
isn't this related to that thread you posted on ODE's? if you have problems with what was done in that thread, or were not sure on how to proceed in that thread then you should have made that known in THAT thread
for $\displaystyle \int \frac {dy}{(y + 3)^2}$ use a substitution, $\displaystyle u = y + 3$
for $\displaystyle \int (x - 1)^2~dx$ expand $\displaystyle (x - 1)^2$ and use the power rule
(and you need to wrap math tags around your LaTeX code to get it to work, that is, you must type [tex]\frac {dy}{(y + 3)^2} = (x - 1)^2~dx[/tex] to get $\displaystyle \frac {dy}{(y + 3)^2} = (x - 1)^2~dx$
It would have been tidier to post this question as part of this thread since it's a follow-up question ....
By makin' $\displaystyle u=x+1$ there's no problem ('cause $\displaystyle du=dx$), so we can use the power rule directly
$\displaystyle \int {(x - 1)^2 \,dx} = \frac{{(x - 1)^3 }}
{3} + k.$
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Solving differential equations without knowing basic integration techniques is quite dangerous.
Brush up some of integration aircraft, from there get in on this.
trying to arrange the equation into this format- dy/dx + P * y = Q
Integrating factor = e ^ int P dx
dy/dx + y/ 2(3-x) = -1/ 2(3-x)
so P = 1/ 2(3-x)
1/ 2(3-x) = 2(3-x)^ -1 so cannot use the normal mathod to integrate, so I use ln instead.
so 1/2 * 1/ (3-x) = 1/2 ln (3-x)
integrating factor = e^ 1/2 in (3-x) and stucked , dunno wat to do next..
correct answer is 1/ square root of (3-x)
Thanks for your help..
Do you know how to use the given integrating factor?
The technique boils down to $\displaystyle y = \frac{1}{\mu} \left( \int \mu Q \, dx + C \right)$, where $\displaystyle \mu = \mu(x)$ is the integrating factor and Q is as you've defined above.
So in your case the solution boils down to
$\displaystyle y = \sqrt{3 - x} \left( -\frac{1}{2}\int \frac{1}{3 - x} \frac{1}{\sqrt{3 - x}} \, dx + C \right) = \sqrt{3 - x} \left( -\frac{1}{2}\int \frac{1}{(3 - x)^{3/2}} \, dx + C \right)$
$\displaystyle = \sqrt{3 - x} \left( -\frac{1}{2}\int (3 - x)^{-3/2} \, dx + C \right)$.
So again, getting the answer boils down to being able to integrate ......