Question

y=x(p²-2p+2)
Please solve for y

Answer 1

Answer to Question #84065 - Math - Differential Equations

Given differential equation

$$\begin{array}{l} y = x \left(p ^ {2} - 2 p + 2\right) \\ \Rightarrow p ^ {2} - 2 p + 2 = y / x \Rightarrow p ^ {2} - 2 p + 1 = \frac {y}{x} - 1 \\ \Rightarrow (p - 1) ^ {2} = \frac {y}{x} - 1 \Rightarrow (p - 1) = \pm \sqrt {\frac {y}{x} - 1} \Rightarrow p = 1 \pm \sqrt {\frac {y}{x} - 1} \\ \Rightarrow \frac {d y}{d x} = 1 \pm \sqrt {\frac {y}{x} - 1} \dots \dots \dots \dots (1) \\ \end{array}$$ $$l e t \frac {y}{x} = t \Rightarrow y = t x \Rightarrow \frac {d y}{d x} = t + x \frac {d t}{d x}$$ $$p u t t i n g \ (1) t + x \frac {d t}{d x} = 1 \pm \sqrt {t - 1}$$ $$\begin{array}{l} \Rightarrow x \frac {d t}{d x} = 1 - t \pm \sqrt {t - 1} \Rightarrow x \frac {d t}{d x} = 1 - t \pm \sqrt {t - 1} \\ \Rightarrow \frac {d t}{1 - t \pm \sqrt {t - 1}} = \frac {d x}{x} \Rightarrow \frac {d t}{\sqrt {t - 1} \left(\pm 1 - \sqrt {t - 1}\right)} = \frac {d x}{x} \\ \end{array}$$

Integrating $\int \frac{dt}{\sqrt{t - 1} \left( \pm 1 - \sqrt{t - 1} \right)} = \int \frac{dx}{x} + \ln c = \ln x + \ln c$ (2)

$$l e t \left(\pm 1 - \sqrt {t - 1}\right) = u \Rightarrow - \frac {d t}{2 \sqrt {t - 1}} = d u \Rightarrow \frac {d t}{\sqrt {t - 1}} = - 2 d u$$

Now (2) becomes $\int \frac{-2du}{u} = \ln x + \ln c \Rightarrow -2\ln u = \ln cx \Rightarrow u^{-2} = cx \Rightarrow u^2 = \frac{1}{cx}$

$$\begin{array}{l} \Rightarrow \left(\pm 1 - \sqrt {t - 1}\right) ^ {2} = \frac {1}{c x} \Rightarrow \pm 1 - \sqrt {t - 1} = \pm \frac {1}{\sqrt {c x}} \Rightarrow \sqrt {t - 1} = \pm 1 \pm \frac {1}{\sqrt {c x}} \\ \Rightarrow \sqrt {\frac {y}{x} - 1} = \pm 1 \pm \frac {1}{\sqrt {c x}} \Rightarrow \frac {y}{x} - 1 = \left(\pm 1 \pm \frac {1}{\sqrt {c x}}\right) ^ {2} \Rightarrow \frac {y}{x} = 1 + \left(\pm 1 \pm \frac {1}{\sqrt {c x}}\right) ^ {2} \Rightarrow y = x \left[ 1 + \left(\pm 1 \pm \frac {1}{\sqrt {c x}}\right) ^ {2} \right] \\ \end{array}$$

hence there are four solutions

$$y = x \left[ 1 + \left(1 + \frac {1}{\sqrt {c x}}\right) ^ {2} \right], \quad y = x \left[ 1 + \left(1 - \frac {1}{\sqrt {c x}}\right) ^ {2} \right], \quad y = x \left[ 1 + \left(- 1 + \frac {1}{\sqrt {c x}}\right) ^ {2} \right], \quad y = x \left[ 1 + \left(- 1 - \frac {1}{\sqrt {c x}}\right) ^ {2} \right]$$

Equivalently we have two solutions $y = x\left[1 + \left(1 + \frac{1}{\sqrt{cx}}\right)^2\right]$ , $y = x\left[1 + \left(1 - \frac{1}{\sqrt{cx}}\right)^2\right]$

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