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[SOLVED] b.2.1.12 Find the general solution

karush

Well-known member
Jan 31, 2012
2,886
Find the general solution
$2y'+y=3t^2$
Rewrite
$y'+\frac{1}{2}y=\frac{3}{2}t^2$
So
$u(t) = exp\left(\int \frac{1}{2} dt \right)=e^{t/2} $


Hopefully so far
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,198
Yep, good so far! How do you continue?
 

karush

Well-known member
Jan 31, 2012
2,886
Find the general solution
$2y'+y=3t^2$
Rewrite
$y'+\frac{1}{2}y=\frac{3}{2}t^2$
So
$u(t) = exp\left(\int \frac{1}{2} dt \right)=e^{t/2} $
then multiply thru
$e^{t/2}y'+\dfrac{e^{t/2}}{2}y=\dfrac{3e^{t/2}}{2}t^2$
which is a product
$(y' \cdot e^{t/2})'=\dfrac{3e^{t/2}}{2}t^2$

$y' \cdot e^{t/2}=\frac{3}{2}\left(\frac{e^{\frac{r}{2}}t^3}{3}+C\right)$
$y'=\dfrac{t^3}{2}+C$

if this is ok then integrate both sides and isolate y' add +C
hard to see if any typos

the integral on this didn't look to good
 
Last edited:

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
854
$(ye^{t/2})’ = \dfrac{3}{2} \cdot t^2 \cdot e^{t/2}$

$\displaystyle ye^{t/2} = \dfrac{3}{2} \int t^2 \cdot e^{t/2} \, dt$

you can use integration by parts on the RHS: $u=t^2$, $dv=e^{t/2} dt$
or tabular integration

you should end up with ...

$y = 3(t^2-4t+8) + Ce^{-t/2}$
 

karush

Well-known member
Jan 31, 2012
2,886
That looks more sensible...

Ummm where did the 2 go?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
854

karush

Well-known member
Jan 31, 2012
2,886
Ok it canceled out