Give the correct answer and solve the given equation dydx=y2−1x2−1,y(2)=2

Given that dydx=y2−1x2−1, y(2)=2 dydx=y2−1x2−1 ⇒dyy2−1=dxx2−1 ⇒∫dyy2−1=∫dxx2−1 ⇒ln⁡(|y−1|)|y+1|=ln⁡(|x−1|)|x+1|+ln⁡c. Now putting y(2)=2 we have ln⁡13=ln⁡13+ln⁡c ⇒13=c13 ⇒c=1 Therefore we have ⇒ln⁡(|y−1|)y+1=ln⁡(|x−1|)|x+1|+ln⁡1 ⇒ln⁡(|y−1|)y+1=ln⁡(|x−1|)|x+1| ⇒|y−1|y+1=|x−1||x+1|

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Bevan Mcdonald

Answered question

2020-11-02

Give the correct answer and solve the given equation

\frac{d y}{d x} = \frac{y^{2} - 1}{x^{2} - 1}, y (2) = 2

Answer & Explanation

Lacey-May Snyder

Skilled2020-11-03Added 88 answers

Given that

\frac{d y}{d x} = \frac{y^{2} - 1}{x^{2} - 1},

y (2) = 2

\frac{d y}{d x} = \frac{y^{2} - 1}{x^{2} - 1}

\Rightarrow \frac{d y}{y^{2} - 1} = \frac{d x}{x^{2} - 1}

\Rightarrow \int \frac{d y}{y^{2} - 1} = \int \frac{d x}{x^{2} - 1}

\Rightarrow \frac{\ln (| y - 1 |)}{| y + 1 |} = \frac{\ln (| x - 1 |)}{| x + 1 |} + \ln c .

Now putting

y (2) = 2

we have

\frac{\ln 1}{3} = \frac{\ln 1}{3} + \ln c

\Rightarrow \frac{1}{3} = c \frac{1}{3}

\Rightarrow c = 1

Therefore we have

\Rightarrow \frac{\ln (| y - 1 |)}{y + 1} = \frac{\ln (| x - 1 |)}{| x + 1 |} + \ln 1

\Rightarrow \frac{\ln (| y - 1 |)}{y + 1} = \frac{\ln (| x - 1 |)}{| x + 1 |}

\Rightarrow \frac{| y - 1 |}{y + 1} = \frac{| x - 1 |}{| x + 1 |}

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