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4th Order Linear ODE

alane1994

Active member
Oct 16, 2012
126
Find the solution of the given initial value problem, and plot its graph. How does the solution behave as \(t\rightarrow\infty\)

\(y^{(4)}-4y'''+4y''=0\)

My work, which coincidentally I believe is incorrect....

From the above differential equation,

\(r^4-4r^3+4r^2=0\)

\(r^2(r-2)^2=0\)

\(r=0,~2\)

Then, would I just input that into a form like this?

\(y=c_1+c_2e^{2t}+c_3te^{2t}\)


I definitely feel as though this isn't correct...
 

chisigma

Well-known member
Feb 13, 2012
1,704
Find the solution of the given initial value problem, and plot its graph. How does the solution behave as \(t\rightarrow\infty\)

\(y^{(4)}-4y'''+4y''=0\)

My work, which coincidentally I believe is incorrect....

From the above differential equation,

\(r^4-4r^3+4r^2=0\)

\(r^2(r-2)^2=0\)

\(r=0,~2\)

Then, would I just input that into a form like this?

\(y=c_1+c_2e^{2t}+c_3te^{2t}\)


I definitely feel as though this isn't correct...
Setting $\displaystyle y^{\ ''}= z$ the ODE becomes...

$\displaystyle z^{\ ''} - 4\ z^{\ '} + 4\ z\ (1)$

... the solution of which is...

$\displaystyle z(t)= c_{1}\ e^{2\ t} + c_{2}\ t\ e^{2\ t}\ (20)$

Now You can solve the ODE...

$\displaystyle y^{ ''} = z\ (3)$

... with two successive integration obtaining...


$\displaystyle y^{\ '} = \int z(t)\ dt\ (4)$


$\displaystyle y = \int y^{ '} (t)\ dt\ (5)$


Kind regards


$\chi$ $\sigma$
 

alane1994

Active member
Oct 16, 2012
126
Thank you very much! That makes quite a bit more sense!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Find the solution of the given initial value problem, and plot its graph. How does the solution behave as \(t\rightarrow\infty\)

\(y^{(4)}-4y'''+4y''=0\)

My work, which coincidentally I believe is incorrect....

From the above differential equation,

\(r^4-4r^3+4r^2=0\)

\(r^2(r-2)^2=0\)

\(r=0,~2\)

Then, would I just input that into a form like this?

\(y=c_1+c_2e^{2t}+c_3te^{2t}\)


I definitely feel as though this isn't correct...
Both of your characteristic roots are repeated, so the general form of your solution would be:

\(\displaystyle y(t)=c_1+c_2t+c_3e^{2t}+c_4te^{2t}\)