MHB Solving 2nd-Order IVP as System of Equations

karush
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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
 
Last edited:
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Since $x_2= x_1$ $x_2'= -5x_2+ sin(t)$. The associated homogeneous equation is $x_2'= \frac{dx}{dt}=-5x_2$ which we can write $\frac{dx_2}{x_2}= -5dt$. Integrating both sides, $ln(x_2)= -5t+ C$. Taking the exponential of both sides, $x_2= e^{-5t+ C}= e^Ce^{-5t}= C'e^{-5t}$.

Since the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x) we look for a solution to the entire equation of the form $x_2= A sin(x)+ B cos(x)$. Then $x_2'= A cos(x)- B sin(x)$ and the equation becomes
$A cos(x)- B sin(x)= -5(A sin(x)+ B cos(x))+ sin(x)$ so that $(A- B)cos(x)+ (5A- B)sin(x)= sin(x)$.

We must have A- B= 0 and 5A- B= 1.
 
Country Boy said:
Since $x_2= x_1$ $x_2'= -5x_2+ sin(t)$. The associated homogeneous equation is $x_2'= \frac{dx}{dt}=-5x_2$ which we can write $\frac{dx_2}{x_2}= -5dt$. Integrating both sides, $ln(x_2)= -5t+ C$. Taking the exponential of both sides, $x_2= e^{-5t+ C}= e^Ce^{-5t}= C'e^{-5t}$.

Since the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x) we look for a solution to the entire equation of the form $x_2= A sin(x)+ B cos(x)$. Then $x_2'= A cos(x)- B sin(x)$ and the equation becomes
$A cos(x)- B sin(x)= -5(A sin(x)+ B cos(x))+ sin(x)$ so that $(A- B)cos(x)+ (5A- B)sin(x)= sin(x)$.

We must have A- B= 0 and 5A- B= 1.
mahalo that helped a lot...
 
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This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...

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