MHB How to Solve Differential Equation with Separate Variables?

[karush]

Spoiler: original

#24 in the spoiler

ov!347.21 nmh{987}
$\dfrac{dy}{dx}=\dfrac{(2¬e^x) }{(3+2y)}\quad y(0)=0$
$\begin{array}{lll}
separate & (3+2y)dy=(2¬e^x)dx \\
integrate &3y + y^2=2x-e^x +c \\
complete the square &y^2+3y+(3/2)^2 =2x-e^x +(3/2)^2+c\\
and &(y+(3/2))^2\\
from &y(0)=0\quad C=1\\
and &y=-\dfrac{3}{2}+\sqrt{2x-e^x+\dfrac{13}{4}}
\end{array}$

ok probably but might have some typos or corrections,
maybe some helpfull suggestion from page in spoiler I am going to try to do all the problems

 

[topsquark]

#24 in the spoiler

$\dfrac{dy}{dx}=\dfrac{(2¬e^x) }{(3+2y)}\quad y(0)=0$
\item\ha$\begin{array}{lll}
\ti{separate varables} & (3+2y)dy=(2¬e^x)dx \\
\ti{integrate} &3y + y^2=2x-e^x +c \\
\ti{complete the square}&y^2+3y+(3/2)^2 =2x-e^x +(3/2)^2+c\\
\ti{and} &(y+(3/2))^2\\
\ti{from} &y(0)=0\ C=1\\
\ti{and} &y=-\dfrac{3}{2}+\sqrt{2x-e^x+\dfrac{13}{4}}
\end{array}$

ok probably but might have some typos or corrections,
maybe some helpfull suggestion from page in spoiler I am going to try to do all the problems

Aside from the [math]\neg[/math] symbol and your incomplete line on "\it{and}" you didn't finish the problem!

But the DEq solution looks good. (But is there any place that you don't have a solution? You never checked.)

-Dan

 

[karush]

Aside from the [math]\neg[/math] symbol and your incomplete line on "\it{and}" you didn't finish the problem!

But the DEq solution looks good. (But is there any place that you don't have a solution? You never checked.)

-Dan

$\dfrac{dy}{dx}=\dfrac{(2¬e^x) }{(3+2y)}\quad y(0)=0$
$\begin{array}{lll}
separate & (3+2y)dy=(2¬e^x)dx \\
integrate &3y + y^2=2x-e^x +c \\
complete \ the \ square &y^2+3y+(3/2)^2 =2x-e^x +(3/2)^2+c\\
reduce &(y+(3/2))^2\\
from \ y(0)=0 &c=1\\
hence &y=-\dfrac{3}{2}+\sqrt{2x-e^x+\dfrac{13}{4}}
\end{array}$

ok I got rid of the format codes from overleaf

the book answer for 2.2.24 was $x=\ln{2}$
but not sure how they got it...

what - are you referring to

 

[topsquark]

$\dfrac{dy}{dx}=\dfrac{(2¬e^x) }{(3+2y)}\quad y(0)=0$
$\begin{array}{lll}
separate & (3+2y)dy=(2¬e^x)dx \\
integrate &3y + y^2=2x-e^x +c \\
complete \ the \ square &y^2+3y+(3/2)^2 =2x-e^x +(3/2)^2+c\\
reduce &(y+(3/2))^2\\
from \ y(0)=0 &c=1\\
hence &y=-\dfrac{3}{2}+\sqrt{2x-e^x+\dfrac{13}{4}}
\end{array}$

ok I got rid of the format codes from overleaf

the book answer for 2.2.24 was $x=\ln{2}$
but not sure how they got it...

what - are you referring to

[math](3+2y)dy=(2¬e^x)dx[/math]

Check the RHS.

C'mon. What is the condition for a function to be at a maximum? y'(x) = 0...

-Dan

 

[karush]

$2¬e^x=0$
$e^x=2\implies \ln(e^x)=\ln{2}\implies x=\ln{2}$
I was expecting something much more exotic:unsure:

Mahalo

 

[topsquark]

$2¬e^x=0$

There it is again. Are you typing this in from a phone?

-Dan

 

[karush]

chromebook I just use the minus sign from the keyboard - in TEX $-$
strange :unsure:

 

[topsquark]

chromebook I just use the minus sign from the keyboard - in TEX $-$
strange :unsure:

Hmmmm... My calculator has two "-" keys. One is subtraction and the other is "negation." I wonder if your chromebook isn't literally using negation?

-Dan

 

[karush]

.

 

[topsquark]

Your link if broken?

-Dan

 

[karush]

https://dl.orangedox.com/wlKD7eKSWiQ79alYD6

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