[SOLVED]DE 20 ty'+(t+1)y=t y(\ln{2}=1, t>0

karush

Well-known member
View attachment 9709
#20 this is the last one of the set
$ty'+(t+1)y=t \quad y(\ln{2}=1,\quad t>0$
rewrite
$y'+\left(\dfrac{t+1}{t}\right)y=1$
$u(t)=e^{\displaystyle\int\dfrac{t+1}{t}dt}=e^{t+\ln |t|}$

well anyway wasn't sure about the (t+1)

Prove It

Well-known member
MHB Math Helper
#20 this is the last one of the set
$ty'+(t+1)y=t \quad y(\ln{2}=1,\quad t>0$
rewrite
$y'+\left(\dfrac{t+1}{t}\right)y=1$
$u(t)=e^{\displaystyle\int\dfrac{t+1}{t}dt}=e^{t+\ln |t|}$

well anyway wasn't sure about the (t+1)
The integrating factor is correct, but again, simplify it so you can integrate.

$\displaystyle \mathrm{e}^{t + \ln{(t)}} = \mathrm{e}^{\ln{(t)}}\,\mathrm{e}^t = t\,\mathrm{e}^t$

karush

Well-known member
ok is that the reason e is used so much???

Ill continue in the morning.....

HallsofIvy

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MHB Math Helper
Is what the reason "e" is used so much? "$$e^x$$" has the nice property that the derivative of $$e^x$$ is just $$e^x$$ again! Also any exponential can be written as "e": $$a^x= e^{ln(a^x)}= e^{x ln(a)}$$.

Those are the reasons "e" is used so much.

karush

Well-known member
The integrating factor is correct, but again, simplify it so you can integrate.

$\displaystyle \mathrm{e}^{t + \ln{(t)}} = \mathrm{e}^{\ln{(t)}}\,\mathrm{e}^t = t\,\mathrm{e}^t$
#20
$\displaystyle u(t)={e}^{t + \ln{(t)}} = {e}^{\ln{(t)}}\,{e}^t = t\,{e}^t$
so
$(t\,{e}^t )ty'+(t\,{e}^t )(t+1)y=(t\,{e}^t )t$
Ok not sure way to rewrite this due to the (t+1) thot could divide by $t^2$

skeeter

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MHB Math Helper
$ty' + (t+1) \cdot y = t$

divide every term by $t$ ...

$y' + \dfrac{t+1}{t} \cdot y = 1$

multiply every term by $te^t$ ...

$te^t \cdot y' + (t+1)e^t \cdot y = te^t$

$(te^t \cdot y)' = te^t$

$\displaystyle te^t \cdot y = \int te^t \, dt$

keep going ...

karush

Well-known member
$ty' + (t+1) \cdot y = t$
divide every term by $t$ ...
$y' + \dfrac{t+1}{t} \cdot y = 1$
multiply every term by $te^t$ ...
$te^t \cdot y' + (t+1)e^t \cdot y = te^t$
$(te^t \cdot y)' = te^t$
$\displaystyle te^t \cdot y = \int te^t \, dt$
keep going ...
IBP
$e^tt-e^t +c$
so
$te^t \cdot y=e^tt-e^t +c$
then
$y=\dfrac{e^tt}{te^t}-\dfrac{e^t }{te^t}+\dfrac{c}{te^t} =1-\dfrac{1}{t}+\dfrac{c}{te^t}$

hopefully

skeeter

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MHB Math Helper
keep going, apply the initial condition to determine $C$

karush

Well-known member
keep going, apply the initial condition to determine $C$
$y=\dfrac{e^tt}{te^t}-\dfrac{e^t }{te^t}+\dfrac{c}{te^t} =1-\dfrac{1}{t}+\dfrac{c}{te^t}=\dfrac{(t-1+(c)e^{-t})}{t}$

$y(\ln 2)=1-\dfrac{1}{\ln 2}+\dfrac{c}{2\ln 2}$
$c=2$

$y=\dfrac{(t-1+2e^{-t})}{t},\quad t\ne 0$

Prove It

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MHB Math Helper
$y=\dfrac{e^tt}{te^t}-\dfrac{e^t }{te^t}+\dfrac{c}{te^t} =1-\dfrac{1}{t}+\dfrac{c}{te^t}=\dfrac{(t-1+(c)e^{-t})}{t}$

$y(\ln 2)=1-\dfrac{1}{\ln 2}+\dfrac{c}{2\ln 2}$
$c=2$
$y=\dfrac{(t-1+2e^{-t})}{t},\quad t\ne 0$