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#### karush

##### Well-known member

- Jan 31, 2012

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- Thread starter karush
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ok yes the 2,2 section exercises are all on separable equationsCould you be clear on what you're struggling with? For instance a) requires first the solution of the DE. Can you identify the kind of DE that it is and suggest a solution approach? (e.g., separable)

I have 10 more problems I want to post here

I'm retired and by myself on this. there is no real need for me me even for me to do this but it certainly has given me a sense of accomplishment

MHB has been a great place for me to spend all the free time I have

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$\dfrac{1}{(4 - y) y} = \dfrac{ 1} {4 y} - \dfrac{1}{4 (y - 4)}$

like this?

do we really need this ugly brown box at the bottom!

- Aug 30, 2012

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You are off by a minus sign.$\dfrac{1}{(4 - y) y} = \dfrac{ 1} {4 y} - \dfrac{1}{4 (y - 4)}$

like this?

do we really need this ugly brown box at the bottom!

\(\displaystyle \dfrac{1}{y(y - 4)} = \dfrac{1}{4y} - \dfrac{1}{4(y - 4)}\)

Watch the negative sign on the 4 - y.

-Dan

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$\displaystyle \dfrac{1}{y(4¬y)} = \dfrac{1}{4y} - \dfrac{1}{4(y - 4)}$ thus if so then $$\dfrac{1}{4} \displaystyle\int \dfrac{1}{y} \, dy- \dfrac{1}{4} \int\dfrac{1}{y-4}\, dy =\dfrac{1}{3} \int t\, dt $$ |

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- Mar 1, 2012

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how about ...

$\displaystyle \dfrac{1}{y(4¬y)} = \dfrac{1}{4y} - \dfrac{1}{4(y - 4)}$

thus

if so then

$$\dfrac{1}{4} \displaystyle\int \dfrac{1}{y} \, dy- \dfrac{1}{4} \int\dfrac{1}{y-4}\, dy

=\dfrac{1}{3} \int t\, dt $$

$$\displaystyle\int \dfrac{1}{y} \, dy- \int\dfrac{1}{y-4}\, dy

=\dfrac{4}{3} \int t\, dt $$

Which ugly box are you talking about?$\dfrac{1}{(4 - y) y} = \dfrac{ 1} {4 y} - \dfrac{1}{4 (y - 4)}$

like this?

do we really need this ugly brown box at the bottom!

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on a tablet the brown footer is a huge retangle with ears on the sidesWhich ugly box are you talking about?

terrifying to stare at

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$$\displaystyle\int \dfrac{1}{y} \, dy- \int\dfrac{1}{y-4}\, dy

=\dfrac{4}{3} \int t\, dt$$

$$\ln |y|-\ln |y-4|=\frac{2t^2}{3}+C$$

ok assume find C from RHS and $y(0) =y_0$

does $y_0=t$??

=\dfrac{4}{3} \int t\, dt$$

$$\ln |y|-\ln |y-4|=\frac{2t^2}{3}+C$$

ok assume find C from RHS and $y(0) =y_0$

does $y_0=t$??

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OK, I am using a Desktop and this ugly box isn't shown there.on a tablet the brown footer is a huge retangle with ears on the sides

terrifying to stare at

y_0 is a constant and t is a variable, so this doesn't make so much sense.$$\displaystyle\int \dfrac{1}{y} \, dy- \int\dfrac{1}{y-4}\, dy

=\dfrac{4}{3} \int t\, dt$$

$$\ln |y|-\ln |y-4|=\frac{2t^2}{3}+C$$

ok assume find C from RHS and $y(0) =y_0$

does $y_0=t$??

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