Find the unique solution to the IVP

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SUMMARY

The unique solution to the initial value problem (IVP) defined by the ordinary differential equation (ODE) $t^3y'' + e^ty' + t^4y = 0$ with conditions $y(1) = 0$ and $y'(1) = 0$ is the trivial solution $y(t) = 0$. Dividing through by $t^4$ simplifies the equation to $\frac{1}{t} y'' + \frac{e^t}{t^4}y' + y = 0$. The discussion confirms that the trivial solution satisfies all requirements of the IVP, although further exploration using exact solution methods is suggested for deeper understanding.

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  • Understanding of ordinary differential equations (ODEs)
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shamieh
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Find the unique solution to the IVP

$t^3y'' + e^ty' + t^4y = 0$ $y(1) = 0$ , $y'(1) = 0$

Should I start out by dividing through by $t^4$

to get

$\frac{1}{t} y" + \frac{e^t}{t^4}y' + y = 0$
 
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I think in this case, you should look for a trivial solution to the given ODE. :D
 
what do you mean? should I plug y(1) = 0
 
Consider the function:

$$y(t)=0$$

Does it satisfy all of the given requirements?
 
Yes?
 
MarkFL said:
Consider the function:

$$y(t)=0$$

Does it satisfy all of the given requirements?

Oh wait.. I need to test this using the exact solution method don't i?

(testing whether it is exact or not)
 
shamieh said:
Oh wait.. I need to test this using the exact solution method don't i?

(testing whether it is exact or not)

That's for first order equations...to be honest, I would not know offhand how to find the general solution to the ODE associated with the IVP here, but I simply noticed that the trivial solution $y(t)=0$ satisfies the IVP. :D
 

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