I needed help with this Differential Equation, below:

$$dy/dt = t + y, \text{ with } y(0) = -1$$

I tried $dy/(t+y) = dt$ and integrated both sides, but it looks like the $u$-substitution does not work out.

  • $\begingroup$ Have you try a change of variable $z=y+t$? $\endgroup$ – Josué Tonelli-Cueto Mar 18 '12 at 11:09
  • $\begingroup$ Why bother changing the variable, when there is a direct formula for solving the equation? If you do not want to apply the direct formula, then the steps in the proof of the formula will give a quick result. en.wikipedia.org/wiki/… $\endgroup$ – Beni Bogosel Mar 18 '12 at 12:12
  • $\begingroup$ I don't know about you, but for me it's easier to remember a method for solving a problem than a formula. $\endgroup$ – Mike Mar 18 '12 at 23:29

This equation is not separable. In other words, you can't write it as $f(y)\;dy=g(t)\;dt$. A differential equation like this can be solved by integrating factors. First, rewrite the equation as:


Now we multiply the equation by an integrating factor so we can use the product rule, $d(uv)=udv+vdu.$ For this problem, that integrating factor would be $e^{-t}$.

$$e^{-t}\frac{dy}{dt}-e^{-t}y=\frac d{dt}(e^{-t}y)=te^{-t}$$

$$e^{-t}y=\int te^{-t}dt=-te^{-t}+\int e^{-t}dt=-te^{-t}-e^{-t}+C$$


For this specific problem, we could also follow Iasafro's suggestion.



As you can see, this substitution resulted in a separable equation, allowing you to integrate both sides.

| cite | improve this answer | |

This is a first order linear differential equation so general solution is given by :

$$y=\frac{\int u(t)\cdot t \,dt +C}{u(t)} ~\text{where}~ u(t)=e^{-\int dt}$$

| cite | improve this answer | |

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.