?

[Spin]

if y_1 and y_2 are the answers of this equation:

(d^4 y/dx^4) + 4 y = f(x)

 

then what's y_2 - y_1?

I used laplace Transform!

 

Thanks in advance

[the tree]

Surely you meant [math]\frac{d^{4}y}{dx^4}4y[/math], not [math]\frac{d^{4}y}{dx^4}+4y[/math]??

 

And what do you mean by "awnsers" anyway?

[NeonBlack]

I wouldn't be so sure about that.

And by answer, he probably means 'What is y?'

[the tree]

Well...

[math]\frac{d^{4}y}{dx^4}4y=\frac{d^{3}y}{dx^3}4=\frac{d^{2}y}{dx^2}0=\frac{dy}{dx}0=0[/math]

So I guess the OPs meaning was correct on that account (somehow), but not knowing the meaning of [math]f(x)[/math], I don't see anything that can be done with it.

[NeonBlack]

f(x) is just an arbitrary function of x.

But since I don't know how to solve 4th order DE's I'm afraid I can't be of much more help.

[Tom Mattson]

if y_1 and y_2 are the answers of this equation:

(d^4 y/dx^4) + 4 y = f(x)

 

then what's y_2 - y_1?

 

OK so your equation is of the form [imath]\mathcal{L}[y]=f(x)[/imath]' date=' where [imath']\mathcal{L}=\frac{d^4}{dx^4}+4[/imath]. Use the fact that [imath]\mathcal{L}[/imath] is linear to find [imath]\mathcal{L}[y_2-y_1][/imath]. You should find that the difference of two solutions to the inhomogeneous equation [imath]\mathcal{L}[y]=f(x)[/imath] is actually a solution to the homogeneous equation [imath]\mathcal{L}[y]=0[/imath]. So your answer will be independent of [imath]f(x)[/imath]. It will of course depend on 4 arbitrary constants of integration.

 

I used laplace Transform!

 

That will work, so continue with it.