I'm pretty sure it's easy but this is the first time I have to solve a differential equation.

Given the following equation to solve:

$ty' + 2y = \sin(t)$.

I have no idea how to start solving it,

can you please explain me detaily?.. thanks in advance.

  • 1
    $\begingroup$ WHat did you try ? $\endgroup$ – Claude Leibovici Jan 25 '14 at 12:34
  • $\begingroup$ @ClaudeLeibovici tried to solve as ordinary equation but it didn't seem to work.. $\endgroup$ – Billie Jan 25 '14 at 12:38
  • $\begingroup$ Ordinary differential equation, do you mean ? $\endgroup$ – Claude Leibovici Jan 25 '14 at 12:43
  • $\begingroup$ @ClaudeLeibovici No, equation without function .. for example, x = 2y $\endgroup$ – Billie Jan 25 '14 at 12:44
  • $\begingroup$ Did you learn about differential equations (they mix functions and their derivatives) ? What you post is not an algebraic equation (you even say it in the title of your post). SO, let us see what you know and we shall continue. $\endgroup$ – Claude Leibovici Jan 25 '14 at 12:47

After multiplication by $t$, we get $$t^2 y'+2 t y=t \sin{t},$$ $$\frac{d}{dt}(t^2 y)=t\sin{t},$$ or $$t^2 y=\int t \sin{t} dt, $$ $$t^2 y=-t \cos{t}+\sin{t}+C.$$ Finally, $$y=-\frac{\cos{t}}{t}+\frac{\sin{t}}{t^2}+\frac{C}{t^2}.$$

  • $\begingroup$ Yes, that was a typo. $\endgroup$ – alans Jan 25 '14 at 13:04

General principle: Usually when you see an expression like $f(t)y'(t)+g(t)y(t)$, a standard idea is to find a function $h$ (called the integrating factor) such that $[f(t)h(t)]'=g(t)h(t)$. Then we note that $[f(t)h(t)y(t)]' = h(t)f(t)y'(t) + h(t)g(t)y(t) = h(t)[f(t)y'(t)+g(t)y(t)]$, and we can use this to simplify the original DE.

In this case, we multiply the equation by a function $h(t)$. This gives $$ h(t) t y'(t) + 2 h(t) y'(t) = h(t) \sin(t) $$ If the function $h$ was chosen such that $[h(t)t]'=2h(t)$, we get $$ [h(t)ty(t)]' = h(t) \sin(t), $$ and we can start solving the problem by integrating, if we know how to integrate the RHS. Remember the constant!

Now the problem is reduced to these parts:

  1. What is a function $h(t)$ such that $[h(t)t]' = 2h(t)$?
  2. What is an integral $F(t)$ of $h(t)\sin(t)$?
  3. Solving $y(t)$ from $h(t) t y(t) = F(t) + C$.
  • $\begingroup$ After reading the comments to the question, I understand this may be too advanced for the OP. Comments are welcome! $\endgroup$ – JiK Jan 25 '14 at 12:54

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