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Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Easy differential equation

The question asks to solve the ODE

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

given that

y(0)=0

.
Is the solution

\displaystyle \frac{1}{2} \ln \left | \frac{x-1}{x+1} \right |

which would be valid for

x\neq \pm 1

OR
Is the solution just

\displaystyle \frac{1}{2} \ln \frac{x-1}{x+1}

which is valid for |x|<1?

OP

Anyone?

Original post by Ano123

The question asks to solve the ODE

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

given that

y(0)=0

.
Is the solution

\displaystyle \frac{1}{2} \ln \left | \frac{x-1}{x+1} \right |

which would be valid for

x\neq \pm 1

OR
Is the solution just

\displaystyle \frac{1}{2} \ln \frac{x-1}{x+1}

which is valid for |x|<1?

\int( \frac{1}{x^2 - 1})dx = -\int(\frac{1}{1-x^2})dx= -artanh(x) +c

So the solution is negative of what you have and it is valid for

|x|<1

(edited 7 years ago)

OP

Original post by NotNotBatman

\int( \frac{1}{x^2 - 1})dx = -\int(\frac{1}{1-x^2})dx= -artanh(x) +c

So the solution is negative of what you have and it is valid for

|x|<1

So in logarithmic form do we still need the absolute value in the log?

Original post by Ano123

So in logarithmic form do we still need the absolute value in the log?

No.

\displaystyle -\frac{1}{2} \ln (\frac{1+x}{1-x}) = -\frac{1}{2}[ln(1+x) - ln(1-x)]

Hopefully you can see that any of x substituted such that

-1<x<1

or

|x|<1

will have a positive argument in the logs, so there is no need for the absolute value.

(edited 7 years ago)

OP

Original post by NotNotBatman

No.

\displaystyle -\frac{1}{2} \ln (\frac{x-1}{x+1}) = -\frac{1}{2}[ln(x-1) - ln(x+1)]

Hopefully you can see that any of x substituted such that

-1<x<1

or

|x|<1

will have a positive argument in the logs, so there is no need for the absolute value.

x=1/2 gives a negative argument in the log.

Original post by Ano123

x=1/2 gives a negative argument in the log.

Typo, now corrected.

OP

Original post by NotNotBatman

Typo, now corrected.

Does

Unparseable latex formula:

\displaystyle \frac{1}{2} \ln \left \frac{1-x}{x+1}

satisfy the ODE for |x|>1? Is this not a valid solution?

(edited 7 years ago)

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