Solving 2nd Order ODEs: y^4 -3y'' -4y = 0

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Homework Help Overview

The discussion revolves around solving a second-order ordinary differential equation (ODE) presented in the form y^4 - 3y'' - 4y = 0. Participants are exploring the classification and potential methods for solving this equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to identify the class of the equation and expresses uncertainty about solving it. Some participants suggest a possible correction to the equation's form, indicating it may be y'''' - 3y'' - 4y = 0. Others mention the complexity of solving the original equation, referencing elliptic integrals. One participant proposes a method involving characteristic equations and exponential solutions.

Discussion Status

The discussion is active, with participants exploring different interpretations of the equation and suggesting various approaches. While some guidance has been offered regarding the characteristic equation method, there is no explicit consensus on the best way to proceed.

Contextual Notes

Participants are navigating potential confusion regarding the equation's form and the implications of its classification as an ODE with constant coefficients. There is mention of the challenges posed by the original equation's complexity.

mj478
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Hi. I am new to differential equations. This is probably pretty easy but I don't quite understand how to do it yet.

The equation is y^4 -3y'' -4y = 0.

I can figure out what class of equation it is. I can write it in the form y'' = F(y), but I am not really sure how to solve it.
 
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Actually I think the problem is y'''' - 3y'' -4y. But I am still not sure what to do.
 
mj478 said:
Actually I think the problem is y'''' - 3y'' -4y. But I am still not sure what to do.

Sure it is y'''' - 3y'' -4y = 0 because dasy to solve.
Solving y^4 - 3y'' -4y =0 is possible, but hard. It involves elliptic integral.
 
This is ODE with constant coefficients..

Suppose that the solution is y=e^rx, than solve it!

i find that r=2, r=-2, r=i, r=-i

which means that

complementary solution of eqn is y=c_1*e^2x + c_2*e^-2x + c_3(sinx) + c_4(cosx)
 

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