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

• MHB
• karush
In summary: 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$(t\,{e}^t karush Gold Member MHB View attachment 9709 #20 this is the last one of the set$ty'+(t+1)y=t \quad y(\ln{2}=1,\quad t>0$rewrite$y'+\left(\dfrac{t+1}{t}\right)y=1u(t)=e^{\displaystyle\int\dfrac{t+1}{t}dt}=e^{t+\ln |t|}$well anyway wasn't sure about the (t+1) karush said: #20 this is the last one of the set$ty'+(t+1)y=t \quad y(\ln{2}=1,\quad t>0$rewrite$y'+\left(\dfrac{t+1}{t}\right)y=1u(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 $ok is that the reason e is used so much? Ill continue in the morning... 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. Prove It said: 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$here are book answersView attachment 9711 #### Attachments • DE.2.1.1.13-20.jpg 23.1 KB · Views: 88$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$keep going ... skeeter said:$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}$hopefully keep going, apply the initial condition to determine$C$skeeter said: keep going, apply the initial condition to determine$Cy=\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}$$y(\ln 2)=1-\dfrac{1}{\ln 2}+\dfrac{c}{2\ln 2} c=2 so the book answer is y=\dfrac{(t-1+2e^{-t})}{t},\quad t\ne 0 karush said: y=\dfrac{e^tt}{te^t}-\dfrac{e^t }{te^t}+\dfrac{c}{te^t} =1-\dfrac{1}{t}+\dfrac{c}{te^t}=\dfrac{(t-1+(c)e^{-t})}{t}$$y(\ln 2)=1-\dfrac{1}{\ln 2}+\dfrac{c}{2\ln 2}c=2$so the book answer is$y=\dfrac{(t-1+2e^{-t})}{t},\quad t\ne 0\$

Come on, surely you can see several common factors that cancel. I don't know why you seem to want to keep around difficult looking terms. The reason it's called "simplifying" is because it literally makes the expressions simpler!

it's not that easy to see

## 1. What does the equation "DE 20 ty'+(t+1)y=t y(\ln{2}=1, t>0" represent?

The equation represents a first-order linear differential equation with a variable coefficient.

## 2. What is the purpose of the "DE 20 ty'+(t+1)y=t y(\ln{2}=1, t>0" equation?

The equation is used to model real-world phenomena that can be described by a linear relationship between a dependent variable and its derivative.

## 3. How do you solve the "DE 20 ty'+(t+1)y=t y(\ln{2}=1, t>0" equation?

The equation can be solved using various methods such as separation of variables, integrating factor, or variation of parameters.

## 4. What are the applications of the "DE 20 ty'+(t+1)y=t y(\ln{2}=1, t>0" equation?

The equation has various applications in physics, engineering, economics, and other fields where the relationship between a variable and its rate of change is important.

## 5. Can the "DE 20 ty'+(t+1)y=t y(\ln{2}=1, t>0" equation be used to predict future values?

Yes, the equation can be used to predict future values by solving for the dependent variable at a given time or by using initial conditions to find a particular solution.

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