- #1

karush

Gold Member

MHB

- 3,269

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$\tiny{2.1.5.1.c}$ source

Change the second-order IVP into a system of equations

$\dfrac{d^2x}{dt^2}+\dfrac{dx}{dt}'+4x=\sin t \quad x(0)=4\quad x'(0)= -3$

ok I presume we can rewrite this as

$u''+u'+4u=\sin t$

Let $x_1=u$ and $x_2=u'$ then $x_1'=x_2$

substituting

$x_2'+x_2+4x=\sin t$

$\begin{array}{lllll}

&let &x_1=u &and &x_2=u'\\

&then &x_1'=x_2 &and &x_2'=u''

\end{array}$

so

$\begin{array}{llll}

x_1'=x_2\\

x_2'=-x_2-4x_1+\sin t

\end{array}$

so far

Change the second-order IVP into a system of equations

$\dfrac{d^2x}{dt^2}+\dfrac{dx}{dt}'+4x=\sin t \quad x(0)=4\quad x'(0)= -3$

ok I presume we can rewrite this as

$u''+u'+4u=\sin t$

Let $x_1=u$ and $x_2=u'$ then $x_1'=x_2$

substituting

$x_2'+x_2+4x=\sin t$

$\begin{array}{lllll}

&let &x_1=u &and &x_2=u'\\

&then &x_1'=x_2 &and &x_2'=u''

\end{array}$

so

$\begin{array}{llll}

x_1'=x_2\\

x_2'=-x_2-4x_1+\sin t

\end{array}$

so far

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