permaneceerc

2021-03-02

Transform the given initial value problem into an initial value problem for two first-order quations $u"+0.25{u}^{\prime }+4u=2\mathrm{cos}\left(3t\right)$, $u\left(0\right)=1,{u}^{\prime }\left(0\right)=-2$

Laaibah Pitt

Let $u={x}_{1}$ and ${u}^{\prime }={x}_{2}$
$⇒{x}_{1}^{\prime }={u}^{\prime }$ and ${x}_{2}^{\prime }=u"$
or ${x}_{1}^{\prime }={x}_{2}$ and ${x}_{2}^{\prime }=u"$
${:}^{\prime }u"+0.25{u}^{\prime }+4u=2\mathrm{cos}\left(3t\right)$
$⇒u"=2\mathrm{cos}\left(3t\right)-0.25{u}^{\prime }-4u$ $\left({:}^{\prime }u={x}_{1},{u}^{\prime }={x}_{2}\right)$
$⇒{x}_{2}^{\prime }=2\mathrm{cos}\left(3t\right)-0.25{x}_{2}-4{x}_{1}$
${x}_{1}^{\prime }={x}_{2}$
${x}_{2}^{\prime }=2\mathrm{cos}\left(3t\right)-4{x}_{1}-0.25{x}_{2}$
or
${x}_{1}^{\prime }=0\cdot {x}_{1}+{x}_{2}$
${x}_{2}^{\prime }=-4{x}_{1}-0.25{x}_{2}+2\mathrm{cos}\left(3t\right)$
${:}^{\prime }u\left(0\right)=1⇒{x}_{1}\left(0\right)=1$ (because $u={x}_{1}$)
${u}^{\prime }\left(0\right)=-2⇒{x}_{2}\left(0\right)=-2$ (because ${u}^{\prime }={x}_{2}$)
Hence the initial value problem for two first-order quations is
${x}_{1}^{\prime }=0\cdot {x}_{1}+1\cdot {x}_{2}$
${x}_{2}^{\prime }=-4{x}_{1}-0.25{x}_{2}+2\mathrm{cos}\left(3t\right)$
${x}_{1}\left(0\right)=1$, ${x}_{2}\left(0\right)=-2$

Jeffrey Jordon

Answer is given below (on video)

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