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$$ty' + ty = 1-y $$

I am having trouble going about this question. I have tried different processes to answer the differential equation, including separable, exact, homogeneous and linear equations. I have ran into a wall through all these processes. Maybe I'm not following the correct procedure... Help! :(

I know the first step would be to divide both sides by $t$, which would give:

$$y' + y = (1-y)/t$$

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$$ty'(t)+ty(t)=1-y(t)\Longleftrightarrow$$ $$y'(t)+\frac{y(t)(t+1)}{t}=\frac{1}{t}\Longleftrightarrow$$


Let $r(t)=\exp\left[\int\frac{t+1}{t}\space\text{d}t\right]=te^t$.

Multiply both sides by $r(t)$:


$$\left(te^t\right)y'(t)+y(t)\left(e^t(t+1)\right)=e^t\Longleftrightarrow$$


Substitute $e^t(t+1)=\frac{\text{d}}{\text{d}t}\left(te^t\right)$:


$$\left(te^t\right)y'(t)+\frac{\text{d}}{\text{d}t}\left(te^t\right)y(t)=e^t\Longleftrightarrow$$


Apply the reverse product rule:


$$\frac{\text{d}}{\text{d}t}\left(ty(t)e^t\right)=e^t\Longleftrightarrow$$ $$\int\frac{\text{d}}{\text{d}t}\left(ty(t)e^t\right)\space\text{d}t=\int e^t\space\text{d}t\Longleftrightarrow$$ $$ty(t)e^t=e^t+\text{C}\Longleftrightarrow$$ $$y(t)=\frac{e^t+\text{C}}{te^t}\Longleftrightarrow$$ $$y(t)=\frac{\text{C}e^{-t}}{t}+\frac{1}{t}$$

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  • $\begingroup$ Thanks! I guess the easiest yet hardest thing as well is actually re-writing the equation to know which technique to use. $\endgroup$ – Alex May 25 '16 at 19:56
  • $\begingroup$ @Alex You're welcome I'm glad that I could help! And yes, just practice. $\endgroup$ – Jan May 25 '16 at 19:58
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Let's try moving the $y$ terms to the LHS:

$$ty'+y(t+1)=1$$

Then divide by $t$:

$y'+y\left(1+\frac{1}{t}\right)=\frac{1}{t}$

Next we want to find the integrating factor: Hint Below

$\mu(t)=e^t t$

Using the integrating factor, you can arrive at the general solution:

$y=\dfrac{e^{-t}(e^t+c)}{t}$

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$ty' +(t+1) y = 1\\ y' +\frac{(t+1)}{t} y = \frac 1t$

Integrating factor $= e^{\int \frac{(t+1)}{t} dt}$

$\int \frac{(t+1)}{t} dt = t + ln t\\ e^{t+lnt} = te^t$

We don't need the arbitrary constant for our integrating factor.

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

Integrate both sides. We have chosen the integrating factor such that the right side integrtes nicely.

$t e^t y = e^t+c\\ y = \frac 1t + \frac ct e^{-t}$

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  • $\begingroup$ Thank you! Definitely helps. $\endgroup$ – Alex May 25 '16 at 19:57

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