Question

Solve for x, dx/dt = x^2-2x+2

Answer 1

@mgeorge20 This is asking to integrate the function

Meaning, what function x has the derivative x^2 - 2x + 2

Answer 2

First we need x on one side and dt on the other

Answer 3

@mgeorge20 So me multiply both sides by dt

and divide both sides by x^2 -2x + 2

Answer 4

\[\frac{ dx }{ dt } = x ^{2} -2x +2 => \frac{ dx }{ x ^{2}-2x +2 }= dt\]

Answer 5

\(\large\color{black}{ \displaystyle dx/dt = x^2-2x+2 }\)

\(\large\color{black}{ \displaystyle \frac{1}{x^2+2x+2}dx/dt =1 }\)

Integrate both sides with respect to t,

\(\large\color{black}{ \displaystyle \int \frac{1}{x^2+2x+2}dx =t+C }\)

\(\large\color{black}{ \displaystyle \int \frac{1}{(x+1)^2+1}dx =t+C }\)

\(\large\color{black}{ \displaystyle \tan^{-1}(x+1)+C =t+C }\)

\(\large\color{black}{ \displaystyle \tan^{-1}(x+1) =t+C }\)

\(\large\color{black}{ \displaystyle x+1 =\tan^{-1}(t+C )}\)

\(\large\color{black}{ \displaystyle x =\tan^{-1}(t+C )-1}\)

Answer 6

integrate both sides

\[\int\limits_{}^{}\frac{ dx }{ x^2-2x+2 }= \int\limits_{}^{}dt\]

Answer 7

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

\[\int\limits_{}^{}\frac{ du }{ a ^{2} + u^2 }\]

Answer 8

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

Answer 9

(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