So I have the following linear differential equation


My first step was to divide through by $t$ to give $$\frac{dy}{dt}-3t^{-1}y=t^3$$

Then to find the integrating factor $e^{\int-3t^{-1}dt}$ which gives $t^{-3}$

I know multiple each side by the integrating factor to get $$\frac{dy/dt}{t^3}-\frac{3y}{t^4}=1$$

Now i cant remember the next step, could anyone continue this problem for me?

  • 2
    $\begingroup$ The LHS is $(y/t^3)'$. $\endgroup$ – David Mitra Jul 11 '15 at 12:32
  • $\begingroup$ @DavidMitra could you explain why? $\endgroup$ – Lauren Bathers Jul 11 '15 at 12:33
  • $\begingroup$ $(e^{\int f} y)'=e^{\int f} y'+y\cdot e^{\int f}\cdot f$, by the product rule. The RHS of this is what you get when you apply your method to ${dy\over dt}+f(t)y=g(t)$ . $\endgroup$ – David Mitra Jul 11 '15 at 12:41
  • $\begingroup$ @DavidMitra oh i see thank you! $\endgroup$ – Lauren Bathers Jul 11 '15 at 13:02

So you're right in obtaining integrating factor of $t^{-3}$ and multiplying through by that. This gives the following:

$$ t^{-3}\frac{dy}{dt} - 3yt^{-4} = 1$$

$$ \frac{d}{dt}(yt^{-3}) = 1$$

Integrating through gives you:

$$ yt^{-3} = \int 1 \ dt $$

Can you finish it from here? Note that multiplying through by an integrating factor makes the differential equation exact which makes it easier to solve.


Although your steps are correct but I suggesting you one more way to find the solution of this differential equation.

The given equation is first order linear differential equation and its standard form is,


Compare the given equation with equation (1) The given differential equation is,


we can also write it as


On comparing equation(1) and (2) we get,

$P(x)=\frac{3}{t}, Q(x)=t^3$

Now the solution of linear differential equation is,

${y}\times{I.F.}=\int{Q(x)}{\times}{I.F.} \; dt\;\;..(3)$

Now, after getting integrating factor(I.F.) put the values of I.F. , P(x) and Q(x) in equation(3)

${y}\times{t^{-3}}=\int{t^{3}}{\times}{t^{-3}} \; dt$

${y}\times{t^{-3}}=\int dt$

${y}\times{t^{-3}}=t+C $

$y=t^{4}+Ct^{3} $

I hope this solution helps you.


It is not difficult to realize that $y(t)=t^4$ is a solution, then by setting $y(t)=t^4+f(t)$ we are left with the differential equation:

$$ t\, f'(t) - 4\, f(t) = 0 $$ that is separable. That gives that every solution is of the form $y(t) = K\,t^4$.


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