MHB 2.1.2 Find the general solution of y'−2y=t^{2}e^{2t}

[karush]

Find the general solution of
$y'−2y=t^{2}e^{2t}$
and use it to determine how solutions behave as
$t \to \infty$

ok presume the first thing to do is to find $u{x}$
$\exp{\displaystyle\int{2} y}=e^{-2} or \dfrac{1}{e^2}$

 

[skeeter]

is the DE
$y' - 2y = t^2e^{2t}$ ?

if so, then the integrating factor is $\mu = e^{\int -2 \, dt} = e^{-2t}$

$e^{-2t} \cdot y' + (-2e^{-2t} \cdot y) = t^2$

$(e^{-2t} \cdot y)' = t^2$

$e^{-2t} \cdot y = \dfrac{t^3}{3} + C$

$y = e^{2t}\left(\dfrac{t^3}{3} + C\right)$

 

[romsek]

do you mean

$$y^\prime - 2y = t^2 e^{2t}$$

if so we have

$$y^\prime + p(t) = q(t)\\
p(t)=-2,~q(t) = t^2e^{2t}$$

The integrating factor is given by

$$\nu(t) = e^{\int p(t)~dt} = e^{-2t}$$

$$e^{-2t}y^\prime - 2 y e^{-2t} = t^2\\

\dfrac{d}{dt}\left(e^{-2t}y\right) = t^2 \\

e^{-2t} y = \dfrac 1 3 t^3 + C\\

y = e^{2t}\left(\dfrac 1 3 t^3 + C\right)

$$

 

[HOI]

y- 2y? That's equal to -y! Do you mean $y'- 2y= t^2e^{2t}$?
You say "the first thing to do is find ux". That makes no sense because there is no "u" or "x" in the problem!

I presume you mean you are looking for an "integrating factor". That would be a function, u(t), such that u(t)y'- 2u(t)y= (u(t)y)'. Since (u(t)y)'= u(t)y'+ u'(t)y we must have u'(t)y= -2u(t)y so that u'(t)= -2u(t). Then u'(t)/u(t)= -2. Integrating, ln(u)= -2t+ C and $u(t)= C'e^{-2t}$ where $C'= e^C$.

Multiplying both sides of $y'- 2y= t^2e^{2t}$ by $e^{-2t}$ (since we can use any integrating factor we can take C'= 1) we get $e^{-2t}y'- 2e^{-2t}y= (e^{-2t}y)'= t^2$.

Integrating both sides $e^{-2t}y= \frac{1}{3}t^3+ C$ so that

$y(t)= Ce^{2t}+ \frac{1}{3}t^3e^{2t}$.

 

[karush]

do you mean

$$y^\prime - 2y = t^2 e^{2t}$$

if so we have

$$y^\prime + p(t) = q(t)\\
p(t)=-2,~q(t) = t^2e^{2t}$$

The integrating factor is given by

$$\nu(t) = e^{\int p(t)~dt} = e^{-2t}$$

$$e^{-2t}y^\prime - 2 y e^{-2t} = t^2\\

\dfrac{d}{dt}\left(e^{-2t}y\right) = t^2 \\

e^{-2t} y = \dfrac 1 3 t^3 + C\\

y = e^{2t}\left(\dfrac 1 3 t^3 + C\right)

$$

yes my bad #2