Solve numerically the IVP

$\frac{d^{2}y}{d{t}^2}+3y\frac{dy}{dt}+6y=-3sint,$ with $,y(0)=1,\frac{dy}{dt}(0)=0$

in the interval $0t60.$ Include the $M-$file in your report.

Is the behavior of the solution significantly different from that of the solution of $(7)?$

Is MATLAB giving any warning message? Comment.

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A Third$-$Order Problem

Consider the third$-$order IVP

$\frac{d^{3}y}{d{t}^3}+3y^2\frac{d^{2}y}{d{t}^2}+6y(\frac{dy}{dt}{)}^2+6\frac{dy}{dt}=-3cost,$ with $,y(0)=1,\frac{dy}{dt}(0)=0,\frac{d^{2}y}{d{t}^2}(0)=0\mathrm{.}5$

Introducing $v=\frac{dy}{dt}$ and $w=\frac{d{y}^2}{d{t}^2}$ we obtain $\frac{dv}{dt}=w$ and $\frac{dw}{dt}=\frac{d^{3}y}{d{t}^3}=-3cost-3y^{2}w-6y{v}^2-6v.$ Moreover,

$v(0)=\frac{dy}{dt}(0)=0$ and $w(0)=\frac{d^{2}y}{d{t}^2}(0)=0\mathrm{.}5.$ Thus $(8)$ is equivalent to

$\frac{dy}{dt}=v,$

$\frac{dv}{dt}=w,$

$\frac{dw}{dt}=-3cost-3y^{2}w-6y{v}^2-6v$

Figure $9$: Time series $y=y(t),v=v(t)=y^{\text{'}}(t),$ and $w=w(t)=y^{\text{'}}(t)($left$),$ and $3$D phase plot $v=y^{\text{'}}$

vs$.y$ vs $w=y^{\text{'}\text{'}}$ for $(8)($rotated with view $([-40,60])).to.=0$;$tf=20$;$yo.=[10$;$60]$;

$a=\mathrm{.}8$;$b=\mathrm{.}01$;$c=\mathrm{.}6$;$d=\mathrm{.}1$;

$[t,y]=$ode$45($@$f,[to.,tf],yo.,[],a,b,c,d)$;

$u1=y($:$,1)$;$u2=y($:$,2)$;$,\%$ y in output has $2$ columns corresponding to u$1$ and u$2$

figure$(1)$;

subplot $(2,1,1)$; plot$($t$,$ u$1,\text{'}$b$-+\text{'})$; ylabel$(\text{'}$u$1\text{'})$;

subplot$(2,1,2)$; plot$($t$,$u$2,\text{'}$ro$-\text{'})$; ylabel$(\text{'}$u$2\text{'})$;

figure$(2)$

plot$($u$1,$u$2)$; axis square; xlabel$(\text{'}$u$\_1\text{'})$; ylabel$(\text{'}$u$\_2\text{'})$; $\%$ plot the phase plot

$\%-----------------------------------------------------------------$

function dydt $=f(t,y,a,b,c,d)$

$u1=y(1)$;$u2=y(2)$;

dydt $=[a^{**}u1-b^{**}u{1}^{**}u2$;$-c^{**}u2+d^{**}u{1}^{**}u2]$;

end