# [SOLVED]b.2.1.12 Find the general solution

#### karush

##### Well-known member
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
Yep, good so far! How do you continue?

#### karush

##### Well-known member
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
$(ye^{t/2})’ = \dfrac{3}{2} \cdot t^2 \cdot e^{t/2}$

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

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
That looks more sensible...

Ummm where did the 2 go?

#### skeeter

##### Well-known member
MHB Math Helper
That looks more sensible...

Ummm where did the 2 go?
do the integration and you’ll find out ...

#### karush

##### Well-known member
Ok it canceled out