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

[mj478]

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.

 

[mj478]

Actually I think the problem is y'''' - 3y'' -4y. But I am still not sure what to do.

 

[JJacquelin]

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.

 

[ertansinan]

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)