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

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:

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


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

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

Multiply both sides by $r(t)$:


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


Apply the reverse product rule:

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

  • $\begingroup$ 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

Let's try moving the $y$ terms to the LHS:


Then divide by $t$:


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:



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

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.

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.