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Solve the following linear ODE $$3t^{2}y'+t^{3}y=cos(t)$$

What i tried

Since this is a linear equation, i used the integrating factor method. First i didide both the LHS and RHS of the equation by $3t^{2}$ so that $y'$ have a coefficient of $1$. Then i find the integrating factor $u(t)=t^{1/3}$. I the multiply the integrating factor of $u(t)$ to both the LHS and RHS of the equation and then integreating both sides of the equation, getting $$yt^{1/3}=\int(1/3)t^{-5/3}cos(t)$$ Im stuck from here onwards though,I tried integration by parts but it seems to lead me to nowhere. Could anyone explain. Thanks

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  • $\begingroup$ The integrating factor is exp $( \int \alpha (t) dt)$, where $\alpha$ is the coefficient of $y$. So I think your integrating factor is wrong. $\endgroup$ – Alfred Yerger Jan 24 '15 at 6:32
  • $\begingroup$ My integrating factor is $e^{1/3ln(t)}$ which simplifies to $t^{1/3}$ $\endgroup$ – ys wong Jan 24 '15 at 6:35
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You have \begin{align} 3t^2\frac{dy}{dt}+t^3y=\cos\left(t\right).\tag{1} \end{align} Now put the equation in standard form as follows: \begin{align} 3t^2\frac{dy}{dt}+t^3y=\cos\left(t\right)\Leftrightarrow \frac{dy}{dt}+\left(\frac{t}{3}\right)y=\frac{\cos\left(t\right)}{3t^2}\tag{2}. \end{align} Notice that the LHS looks a lot like the product rule of differentiation. Therefore, we must find an integrating factor $\mu\left(t\right)$: \begin{align} &\mu\left(t\right) = \exp\left(\int \frac{t}{3}\:dt\right)=e^{t^2/6},\tag{3} \\ \therefore e^{t^2/6}\frac{dy}{dx}+\left(\frac{t}{3}\right)&e^{t^2/6}y=\frac{d}{dt}\left[\left(e^{t^2/6}\right)\left(y\right)\right]=\frac{e^{t^2/6}\cos\left(t\right)}{3t^2} \tag{4} \\ &\implies \int\frac{d}{dt}\left[\left(e^{t^2/6}\right)\left(y\right)\right]\:dt=\int\frac{e^{t^2/6}\cos\left(t\right)}{3t^2}\:dt\tag{5}\\ &\implies y\left(t\right)=e^{-t^2/6}\int\frac{e^{t^2/6}\cos\left(t\right)}{3t^2}\:dt.\tag{6} \end{align} Now solve for the integral (it's not too bad).

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  • $\begingroup$ Is it through integration by parts? $\endgroup$ – ys wong Jan 24 '15 at 6:48
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    $\begingroup$ I am not too sure that "it's not too bad". $\endgroup$ – Claude Leibovici Jan 24 '15 at 7:49
  • $\begingroup$ Actually this is the difficult part of the whole question. $\endgroup$ – ys wong Jan 24 '15 at 7:56
  • $\begingroup$ Could anyone help me with this? This is where im actually stuck at. $\endgroup$ – ys wong Jan 24 '15 at 14:45

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