\(\displaystyle{m}^{{2}}-{4}{m}+{13}={0}\to{m}={2}\pm{3}{i}\)

Thus the general solution is:

\(\displaystyle{y}={e}^{{{2}{x}}}{\left({A}{\cos{{\left({3}{x}\right)}}}+{B}{\sin{{\left({3}{x}\right)}}}\right)}\)

movingsupplyw1

Answered 2021-12-18
Author has **5612** answers

What do you mean as a x missing case ? This is a second order linear differential equation, not a first order one.

Assume a solution is proportional to \(\displaystyle{e}^{{\lambda{x}}}\) for some constant \(\displaystyle\lambda\). Substitute \(\displaystyle{y}{\left({x}\right)}={e}^{{\lambda{x}}}\) into the ODE :

\(\displaystyle\lambda^{{2}}{e}^{{\lambda{x}}}-{4}\lambda{e}^{{\lambda{x}}}+{13}{e}^{{\lambda{x}}}={0}\Rightarrow{e}^{{\lambda{x}}}{\left(\lambda^{{2}}-{4}\lambda+{13}\right)}\)

\(\displaystyle={0}\Rightarrow\lambda={2}\pm{3}{i}\)

The roots \(\displaystyle\lambda={2}\pm{3}{i}\) give \(\displaystyle{y}_{{1}}{\left({x}\right)}={c}_{{1}}{e}^{{{\left({2}+{3}{i}\right)}{x}}}\) and \(\displaystyle{y}_{{2}}{\left({x}\right)}={e}^{{{\left({2}-{3}{i}\right)}{x}}}\). The general solution is the sum of the above equations :

\(\displaystyle{y}{\left({x}\right)}={y}_{{1}}{\left({x}\right)}+{y}_{{2}}{\left({x}\right)}\Rightarrow{y}_{{g{{\left({x}\right)}}}}={c}_{{1}}{e}^{{{\left({2}+{3}{i}\right)}{x}}}+{c}_{{2}}{e}^{{{\left({2}-{3}{i}\right)}{x}}}\)

Now, by applying Euler's Identity

\(\displaystyle{e}^{{{a}+{i}{b}}}={e}^{{a}}{\cos{{b}}}+{i}{e}^{{a}}{\sin{{b}}}\) and regrouping terms, you can yield a final form for the solution :

\(\displaystyle{y}{\left({x}\right)}={c}_{{1}}{e}^{{{2}{x}}}{\cos{{\left({3}{x}\right)}}}+{c}_{{2}}{e}^{{{2}{x}}}{\sin{{\left({3}{x}\right)}}}\)

Assume a solution is proportional to \(\displaystyle{e}^{{\lambda{x}}}\) for some constant \(\displaystyle\lambda\). Substitute \(\displaystyle{y}{\left({x}\right)}={e}^{{\lambda{x}}}\) into the ODE :

\(\displaystyle\lambda^{{2}}{e}^{{\lambda{x}}}-{4}\lambda{e}^{{\lambda{x}}}+{13}{e}^{{\lambda{x}}}={0}\Rightarrow{e}^{{\lambda{x}}}{\left(\lambda^{{2}}-{4}\lambda+{13}\right)}\)

\(\displaystyle={0}\Rightarrow\lambda={2}\pm{3}{i}\)

The roots \(\displaystyle\lambda={2}\pm{3}{i}\) give \(\displaystyle{y}_{{1}}{\left({x}\right)}={c}_{{1}}{e}^{{{\left({2}+{3}{i}\right)}{x}}}\) and \(\displaystyle{y}_{{2}}{\left({x}\right)}={e}^{{{\left({2}-{3}{i}\right)}{x}}}\). The general solution is the sum of the above equations :

\(\displaystyle{y}{\left({x}\right)}={y}_{{1}}{\left({x}\right)}+{y}_{{2}}{\left({x}\right)}\Rightarrow{y}_{{g{{\left({x}\right)}}}}={c}_{{1}}{e}^{{{\left({2}+{3}{i}\right)}{x}}}+{c}_{{2}}{e}^{{{\left({2}-{3}{i}\right)}{x}}}\)

Now, by applying Euler's Identity

\(\displaystyle{e}^{{{a}+{i}{b}}}={e}^{{a}}{\cos{{b}}}+{i}{e}^{{a}}{\sin{{b}}}\) and regrouping terms, you can yield a final form for the solution :

\(\displaystyle{y}{\left({x}\right)}={c}_{{1}}{e}^{{{2}{x}}}{\cos{{\left({3}{x}\right)}}}+{c}_{{2}}{e}^{{{2}{x}}}{\sin{{\left({3}{x}\right)}}}\)

RizerMix

Answered 2021-12-29
Author has **9349** answers

In order to make your problem solvable in the manner in which you seek I think a substitution of \(y=e^{2t}w\) is required. Here w is the new dependent variable. We calculate,

\(y'=e^{2t}(w'+2w)\)

\(y''=e^{2t}(w''+4w'+4w)\)

substituting into \(t''+4y'+9y=0\) yields:

\(e^{2t}(w''+9w)=0\)

hence solve the much easier problem \(w''+9w=0\) by a technique like the one you mention. Personally, I prefer the following trick \(w''=v\frac{dv}{dw}\ where\ v=\frac{dw}{dt}\)

Thus, \(w''+9w=0\) yields

\(v\frac{dv}{dw}+9w=0\)

or

\(vdv+9wdw=d(v^2/2+9w^2/2)=0\)

hence \(v^2+9w^2=9C^2\) is the solution. Solve for \(v=\pm\sqrt{9C^2-9w^2}\) But \(v=\frac{dw}{dt}\) hence we reduce to the quadrature:

\(t=\pm\int\frac{dw}{\sqrt{(C^2-9w^2}}=\frac{\pm}{3|C|}\int\frac{dw}{\sqrt{1-w^2/C^2}}\)

Let \(u=w/C\) hence \(du=dw/C\) and

\(\int\frac{du}{\sqrt{1-u^2}}=\sin^{-1}(u)+c_1\). Consequently,

\(t=\frac{\pm C}{3|C|}(\sin^{-1}(u)+c_1)\)

Or,

\(\sin^{-1}(w/C)=\pm3t-c_1\)

Yielding,

\(w=C\sin(\pm3t-c_1)=A\sin(3t+\phi)\)

Since \(y=e^{2t}w\) we conclude,

\(y=Ae^{2t}\sin(3t+\phi)=c_2e^{2t}\sin(3t)+c_3e^{2t}\cos(3t)\)

asked 2021-11-22

asked 2021-11-23

Solve th second order linear equations:

\(\displaystyle{y}{''}+{4}{y}={\cos{{2}}}{x}\)

\(\displaystyle{y}{''}+{4}{y}={\cos{{2}}}{x}\)

asked 2021-11-21

I need to find a second order linear homogeneous equation with constant coefficients that has the given function as a solution

Queston a) \(\displaystyle{x}{e}^{{-{3}{x}}}\)

Question b) \(\displaystyle{e}^{{{3}{x}}}{\sin{{x}}}\)

Queston a) \(\displaystyle{x}{e}^{{-{3}{x}}}\)

Question b) \(\displaystyle{e}^{{{3}{x}}}{\sin{{x}}}\)

asked 2021-12-10

I am trying to solve the following:

\(\displaystyle{y}{''}+{4}{y}'={\tan{{\left({t}\right)}}}\)

I have used the method of variation of parameters. Currently I am at a point in the equation where I have this:

\(\displaystyle{u}_{{1}}=\int{\frac{{{\tan{{t}}}{\cos{{2}}}{t}}}{{{2}}}}\)

I am stuck here

\(\displaystyle{y}{''}+{4}{y}'={\tan{{\left({t}\right)}}}\)

I have used the method of variation of parameters. Currently I am at a point in the equation where I have this:

\(\displaystyle{u}_{{1}}=\int{\frac{{{\tan{{t}}}{\cos{{2}}}{t}}}{{{2}}}}\)

I am stuck here

asked 2021-11-21

How can the following second-order linear equation be converted into a first-order linear equation?

This is our second-order equation:

\(\displaystyle{y}{''}-{2}{y}'+{2}{y}={e}^{{{2}{t}}}{\sin{{t}}}\)

This is our second-order equation:

\(\displaystyle{y}{''}-{2}{y}'+{2}{y}={e}^{{{2}{t}}}{\sin{{t}}}\)

asked 2021-12-18

Let \(\displaystyle{y}{\left({t}\right)}\) be a nontrivial solution for the second order differential equation

\(\displaystyle\ddot{{{x}}}+{a}{\left({t}\right)}\dot{{{x}}}+{b}{\left({t}\right)}{x}={0}\)

to determine a solution that is linearly independent from y we set \(\displaystyle{z}{\left({t}\right)}={y}{\left({t}\right)}{v}{\left({t}\right)}\)

Show that this leads to a first order differential equation for \(\displaystyle\dot{{{v}}}={w}\)

\(\displaystyle\ddot{{{x}}}+{a}{\left({t}\right)}\dot{{{x}}}+{b}{\left({t}\right)}{x}={0}\)

to determine a solution that is linearly independent from y we set \(\displaystyle{z}{\left({t}\right)}={y}{\left({t}\right)}{v}{\left({t}\right)}\)

Show that this leads to a first order differential equation for \(\displaystyle\dot{{{v}}}={w}\)

asked 2021-11-23

I have the following differential equation:

\(\displaystyle{y}{''}+{y}={\cos{{\left({t}\right)}}}{\cos{{\left({2}{t}\right)}}}\)

Maybe something can be done to \(\displaystyle{\cos{{\left({t}\right)}}}{\cos{{\left({2}{t}\right)}}}\) to make it easier to solve. Any ideas?

\(\displaystyle{y}{''}+{y}={\cos{{\left({t}\right)}}}{\cos{{\left({2}{t}\right)}}}\)

Maybe something can be done to \(\displaystyle{\cos{{\left({t}\right)}}}{\cos{{\left({2}{t}\right)}}}\) to make it easier to solve. Any ideas?