- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

$2y'+y=3t^2$

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

- Thread starter karush
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 2,886

$2y'+y=3t^2$

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

- Admin
- #2

- Jan 26, 2012

- 4,198

Yep, good so far! How do you continue?

- Thread starter
- #3

- 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

$2y'+y=3t^2$

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:

- Mar 1, 2012

- 854

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

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 ...

$y = 3(t^2-4t+8) + Ce^{-t/2}$

- Thread starter
- #5

- Jan 31, 2012

- 2,886

That looks more sensible...

Ummm where did the 2 go?

Ummm where did the 2 go?

- Mar 1, 2012

- 854

do the integration and you’ll find out ...That looks more sensible...

Ummm where did the 2 go?

- Thread starter
- #7

- Jan 31, 2012

- 2,886

Ok it canceled out