Solve 4th Order Linear ODE and Plot Graph: t→∞

[alane1994]

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]

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]

Thank you very much! That makes quite a bit more sense!

 

[MarkFL]

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:

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

 

[blue_raver22]

I would like to provide a more detailed and accurate response to this content. Firstly, the given differential equation is a 4th order linear ODE, which means it is an equation that involves the 4th derivative of the dependent variable, y, and it is linear, meaning that the dependent variable and its derivatives appear only in the first power. The general form of a 4th order linear ODE is:

\(a_4y^{(4)}+a_3y^{(3)}+a_2y''+a_1y'+a_0y=g(t)\)

where \(a_i\) are constants and \(g(t)\) is a function of the independent variable, t. In this case, the given equation has no function on the right-hand side, so \(g(t)=0\).

To solve this differential equation, we can use the method of characteristic equation. We first write the characteristic equation as:

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

Then, we can factor out \(r^2\) to get:

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

which gives us two roots, \(r=0\) and \(r=2\). This means that the general solution of the given ODE is:

\(y=c_1+c_2e^{2t}+c_3te^{2t}+c_4t^2e^{2t}\)

where \(c_i\) are constants to be determined by the initial conditions. This solution is obtained by using the method of undetermined coefficients, where we assume that the solution has the form of the right-hand side of the equation and then solve for the constants.

To plot the graph of this solution, we can use a graphing software or hand-draw it. As \(t\rightarrow\infty\), the solution behaves as \(y\rightarrow c_1+c_2e^{2t}\). This means that the solution approaches a constant value, which is \(c_1\) as \(t\rightarrow\infty\). The other terms in the solution, \(c_3te^{2t}+c_4t^2e^{2t}\), will become negligible compared to the exponential term as \(t\) gets larger, so we can ignore them in the long-term behavior of the solution.