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## Homework Statement:

- Solve dy/dx = (y^2 - 1)/(x^2 - 1), y(2) = 2

## Relevant Equations:

- dy/dx = (y^2 - 1)/(x^2 - 1), y(2) = 2

This is my attempt:

[tex]\frac{dy}{dx} = \frac{y^2 - 1}{x^2 - 1}

\\ \int \frac{dy}{y^2 - 1} = \int \frac{dx}{x^2 - 1}

\\ \ln \left| \frac{y-1}{y+1} \right| + C_1 = \ln \left| \frac{x-1}{x+1} \right| + C_2

\\ \ln \left| \frac{y-1}{y+1} \right| = \ln \left| \frac{x-1}{x+1} \right| + C [/tex]

Since y(2) = 2,

[tex]\ln \left| \frac{1}{3} \right| = \ln \left| \frac{1}{3} \right| + C

\\ \therefore C = 0[/tex]

So,

[tex]\ln \left| \frac{y-1}{y+1} \right| = \ln \left| \frac{x-1}{x+1} \right|

\\ \left| \frac{y-1}{y+1} \right| = \left| \frac{x-1}{x+1} \right|

\\ \frac{y-1}{y+1} = \pm \frac{x-1}{x+1}

\\y = x, y = \frac{1}{x}[/tex]

Is this ok, or did I make any mistake?

[tex]\frac{dy}{dx} = \frac{y^2 - 1}{x^2 - 1}

\\ \int \frac{dy}{y^2 - 1} = \int \frac{dx}{x^2 - 1}

\\ \ln \left| \frac{y-1}{y+1} \right| + C_1 = \ln \left| \frac{x-1}{x+1} \right| + C_2

\\ \ln \left| \frac{y-1}{y+1} \right| = \ln \left| \frac{x-1}{x+1} \right| + C [/tex]

Since y(2) = 2,

[tex]\ln \left| \frac{1}{3} \right| = \ln \left| \frac{1}{3} \right| + C

\\ \therefore C = 0[/tex]

So,

[tex]\ln \left| \frac{y-1}{y+1} \right| = \ln \left| \frac{x-1}{x+1} \right|

\\ \left| \frac{y-1}{y+1} \right| = \left| \frac{x-1}{x+1} \right|

\\ \frac{y-1}{y+1} = \pm \frac{x-1}{x+1}

\\y = x, y = \frac{1}{x}[/tex]

Is this ok, or did I make any mistake?

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