1. ## differential and proof

Hey guys!

How would you solve the following:

1) If $\displaystyle y = e^{-kt}(Acoshqt +Bsinhqt)$ [A, B, k, q are constants]

show that $\displaystyle \frac{d^2y}{dt^2} + 2k\frac{dy}{dt} + (k^2 - q^2)y=0$

2) Find $\displaystyle \frac{dy}{dx}$ when $\displaystyle y = tanh^{-1}\left(\frac{2x}{1+x^2}\right)$

Thanks.

2) Find $\displaystyle \frac{dy}{dx}$ when $\displaystyle y \,= \,\text{tanh}^{-1}\left(\frac{2x}{1+x^2}\right)$

We have: .$\displaystyle \text{tanh }y \:=\:\frac{2x}{1+x^2}$

From the defintion of $\displaystyle \text{tanh }y$, we have: .$\displaystyle \frac{e^y - e^{-y}}{e^y + e^{-y}} \:=\:\frac{2x}{1+x^2}$

Multiply the left fraction by $\displaystyle \frac{e^y}{e^y}\!:\;\;\frac{e^{2y} - 1}{e^{2y} + 1} \:=\:\frac{2x}{1 + x^2}$

. . $\displaystyle (1+x^2)(e^{2y} - 1) \:=\:2x(e^{2y} + 1)\quad\Rightarrow\quad (1+x^2)e^{2y} - (1 + x^2) \:=\:2xe^{2y} + 2x$

. . $\displaystyle (1 + x^2)e^{2y} - 2xe^{2y} \:=\:2x + (1 + x^2)$

. . $\displaystyle (x^2 - 2x + 1)e^{2y}\:=\:x^2 + 2x + 1$

. . $\displaystyle e^{2y} \:=\:\frac{x^2+2x+1}{x^2-2x+1} \:=\:\frac{(x+1)^2}{(x-1)^2} \:=\:\left(\frac{x+1}{x-1}\right)^2$

. . $\displaystyle 2y \:=\:\ln\left(\frac{x+1}{x-1}\right)^2 \:=\:2\ln\left(\frac{x+1}{x-1}\right)$

. . $\displaystyle y \;=\;\ln\left(\frac{x+1}{x-1}\right) \;= \;\ln(x+1) - \ln(x-1)$

Therefore: .$\displaystyle \frac{dy}{dx} \;=\;\frac{1}{x+1} - \frac{1}{x - 1}\;=\;\frac{-2}{x^2-1} \;=\;\frac{2}{1-x^2}$

3. thanks!

can any one help with question 1?

Hey guys!

How would you solve the following:

1) If $\displaystyle y = e^{-kt}(Acoshqt +Bsinhqt)$ [A, B, k, q are constants]

show that $\displaystyle \frac{d^2y}{dt^2} + 2k\frac{dy}{dt} + (k^2 - q^2)y=0$
$\displaystyle y = e^{-kt}(A \cosh(qt) +B\sinh(qt))$

$\displaystyle y' = -k e^{-kt}(A \cosh(qt) +B\sinh(qt)) +e^{-kt}q (A \sinh(qt) +B\cosh(qt))$

.....$\displaystyle = -k y +e^{-kt}q (A \sinh(qt) +B\cosh(qt))$

So:

$\displaystyle y'' = -k y' -k e^{-kt}q (A \sinh(qt)$$\displaystyle +B\cosh(qt))+e^{-kt}q^2 (A \cosh(qt) +B\sinh(qt)) hence: \displaystyle y'' = -k y' -k e^{-kt}q (A \sinh(qt)$$\displaystyle +B\cosh(qt))+q^2y$

Now rework the middle term:

$\displaystyle y'' = -k y' -k (y'+ky) +q^2y$

Which rearranges into the required differential equation.

RonL

5. Originally Posted by CaptainBlack
$\displaystyle y' = -k e^{-kt}(A \cosh(qt) +B\sinh(qt)) +e^{-kt}q (A \sinh(qt) +B\cosh(qt))$

.....$\displaystyle = -k y +e^{-kt}q (A \sinh(qt) +B\cosh(qt))$

So:

$\displaystyle y'' = -k y' -k e^{-kt}q (A \sinh(qt)$$\displaystyle +B\cosh(qt))+e^{-kt}q^2 (A \cosh(qt) +B\sinh(qt)) hence: \displaystyle y'' = -k y' -k e^{-kt}q (A \sinh(qt)$$\displaystyle +B\cosh(qt))+q^2y$

Now rework the middle term:

$\displaystyle y'' = -k y' -k (y'+ky) +q^2y$

Which rearranges into the required differential equation.

RonL
hey captain!

what do you do on the second line starting y' and also line y''.

thanks

6. Originally Posted by CaptainBlack
$\displaystyle y = e^{-kt}(A \cosh(qt) +B\sinh(qt))$

$\displaystyle y' = -k e^{-kt}(A \cosh(qt) +B\sinh(qt)) +e^{-kt}q (A \sinh(qt) +B\cosh(qt))$

.....$\displaystyle = -k y +e^{-kt}q (A \sinh(qt) +B\cosh(qt))$

So:

$\displaystyle y'' = -k y' -k e^{-kt}q (A \sinh(qt)$$\displaystyle +B\cosh(qt))+e^{-kt}q^2 (A \cosh(qt) +B\sinh(qt)) hence: \displaystyle y'' = -k y' -k e^{-kt}q (A \sinh(qt)$$\displaystyle +B\cosh(qt))+q^2y$

Now rework the middle term:

$\displaystyle y'' = -k y' -k (y'+ky) +q^2y$

Which rearranges into the required differential equation.

RonL
hey captain!

what do you do on the second line starting y' and also line y''.

thanks

The first line where y' appears is using the product rule on the previous line (the expression for y). Where the $\displaystyle e^{-kt}$ is one term and $\displaystyle A \cosh(qt)+B \sinh(qt)$ is the other term. Then you need to know that $\displaystyle (d/dx) \cosh(x)=\sinh(x)$ and $\displaystyle (d/dx) \sinh(x)=\cosh(x)$.

Similarly where y'' appears this is the derivative of y', the first term is the derivative of the first term in y', and the remaining terms come from applying the product rule to the second term in the expression for y'.

RonL

7. Originally Posted by CaptainBlack

$\displaystyle y'' = -k y' -k (y'+ky) +q^2y$

Which rearranges into the required differential equation.

RonL
thank you

8. Originally Posted by CaptainBlack

Now rework the middle term:

$\displaystyle y'' = -k y' -k (y'+ky) +q^2y$
How do you get -k(y' +ky) in the quoted text?

Thank you

How do you get -k(y' +ky) in the quoted text?

Thank you
From the top of that post we have:

y' = -k y +e^{-kt}q (A sinh(qt) +B cosh(qt))

y' + ky = e^{-kt}q (A sinh(qt) +B cosh(qt))

which should do the job.

RonL