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So I have the following linear differential equation

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

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?

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    $\begingroup$ 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
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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.

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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,

$\frac{dy}{dx}+P(x)=Q(x)dx\;\;..(1)$

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

$t−3dydt−3yt−4=1$

we can also write it as

$\frac{dy}{dx}+(\frac{3}{t})y=t^3\;\;..(2)$

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.

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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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