anonymous
  • anonymous
Solve for x, dx/dt = x^2-2x+2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@mgeorge20 This is asking to integrate the function Meaning, what function x has the derivative x^2 - 2x + 2
anonymous
  • anonymous
First we need x on one side and dt on the other
anonymous
  • anonymous
@mgeorge20 So me multiply both sides by dt and divide both sides by x^2 -2x + 2

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anonymous
  • anonymous
\[\frac{ dx }{ dt } = x ^{2} -2x +2 => \frac{ dx }{ x ^{2}-2x +2 }= dt\]
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle dx/dt = x^2-2x+2 }\) \(\large\color{black}{ \displaystyle \frac{1}{x^2+2x+2}dx/dt =1 }\) Integrate both sides with respect to t, \(\large\color{black}{ \displaystyle \int \frac{1}{x^2+2x+2}dx =t+C }\) \(\large\color{black}{ \displaystyle \int \frac{1}{(x+1)^2+1}dx =t+C }\) \(\large\color{black}{ \displaystyle \tan^{-1}(x+1)+C =t+C }\) \(\large\color{black}{ \displaystyle \tan^{-1}(x+1) =t+C }\) \(\large\color{black}{ \displaystyle x+1 =\tan^{-1}(t+C )}\) \(\large\color{black}{ \displaystyle x =\tan^{-1}(t+C )-1}\)
anonymous
  • anonymous
integrate both sides \[\int\limits_{}^{}\frac{ dx }{ x^2-2x+2 }= \int\limits_{}^{}dt\]
anonymous
  • anonymous
the only way we can integrate the above function (which I re-wrote, I don't know why) \[\int\limits_{}^{}\frac{ 1 }{ x ^{2}-2x +2}dx\] we need to get it in the form of \[\int\limits_{}^{}\frac{ du }{ a ^{2} + u^2 }\]
anonymous
  • anonymous
Complete the square \[x^2-2x+2 => x^2 -2x + (0.5*-2)^1 - (0.5*-2)^2 + 2\] \[x^2 -2x + (-1)^2 - (-1)^2 + 2 => x^2 -2x + 1 -1 + 2\] \[(x - 1)^2 - 1 + 2\] \[(x - 1)^2 + 1^2\] a = 1 u = x - 1
anonymous
  • anonymous
(rule # 17) \[\int\limits_{}^{} \frac{ du }{ a^2 + u^2} = \frac{ 1 }{ a } \arctan \frac{ u }{ a } + C\] \[\int\limits_{}^{}\frac{ dx }{ (x-1)^2 + 1^2}= \int\limits_{}^{}dt\] \[\frac{ 1 }{ 1 }\arctan \frac{ x-1 }{ 1 } + c_1 = t + c_2\] arctan (x-1) = (t + k) tan arctan(x-1) = tan (t + k) x -1 = tan (t + k) Final answer x = tan(t + k) + 1 or x = tan(t + c) + 1

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