-242.7x.13 Solve the DE \dfrac{dy}{dx}=7x^6e^{-y}

[karush]

$\tiny{242.7x.13}$
1000
$\textsf{Solve the differential equation}$
\begin{align*}\displaystyle
\frac{dy}{dx}&=7x^6e^{-y}\\
dy&=7x^6e^{-y} \, dx \\
e^y dy&=7x^6 dx
\end{align*}
$\textit{just seeing if going the right direction on this...}$

 

[MarkFL]

I've moved the thread to our "Differential Equations" forum. ;)

Yes, you have been given a separable ODE, and you have separated the variables correctly, now you are ready to integrate. :D

 

[karush]

$\tiny{242.7x.13}$
$\textsf{Solve the differential equation}$
\begin{align*}\displaystyle
\frac{dy}{dx}&=7x^6e^{-y}\\
dy&=7x^6e^{-y} \, dx \\
e^y dy&=7x^6 dx
\end{align*}
$\textit{Integrate both sides}$
\begin{align*}\displaystyle
\int e^y dy &=\int 7x^6 dx\\
e^y&=x^7
\end{align*}
$\textit{Isolate y}$
\begin{align*}\displaystyle
ln (e)^y&=ln\left({x}\right)^7 \\
y&=7ln(x)
\end{align*}
$\textit{Is it this ??}$

 

[MarkFL]

$\tiny{242.7x.13}$
$\textsf{Solve the differential equation}$
\begin{align*}\displaystyle
\frac{dy}{dx}&=7x^6e^{-y}\\
dy&=7x^6e^{-y} \, dx \\
e^y dy&=7x^6 dx
\end{align*}
$\textit{Integrate both sides}$
\begin{align*}\displaystyle
\int e^y dy &=\int 7x^6 dx\\
e^y&=x^7
\end{align*}
$\textit{Isolate y}$
\begin{align*}\displaystyle
ln (e)^y&=ln\left({x}\right)^7 \\
y&=7ln(x)
\end{align*}
$\textit{Is it this ??}$

You forgot the constant of integration...but otherwise, you did fine. :D

 

[karush]

\begin{align*}\displaystyle
ln (e)^y&=ln\left({x}\right)^7 \\
y&=7ln(x)+C
\end{align*}

 

[MarkFL]

\begin{align*}\displaystyle
ln (e)^y&=ln\left({x}\right)^7 \\
y&=7ln(x)+C
\end{align*}
$\text{easy to forget our C friend...}$

You can't simply "tack it on" there...it has to be there before you take logs to solve for $y$...;)

\(\displaystyle e^y=x^7+C\)

\(\displaystyle y=\ln\left(x^7+C\right)\)

 

[Theia]

One comment on notation too:

$\textit{Isolate y}$
\begin{align*}\displaystyle
ln (e)^y&=ln\left({x}\right)^7 \\
y&=7ln(x)
\end{align*}
$\textit{Is it this ??}$

The way you have written this can be confusing, since \(\displaystyle \ln (a^b) = b \ln(a)\). But at least I read \(\displaystyle \ln (a)^b\) as \(\displaystyle (\ln a)^b\). And for this we know \(\displaystyle (\ln a)^b \ne b \ln (a)\).

 

[karush]

One comment on notation too:

The way you have written this can be confusing, since \(\displaystyle \ln (a^b) = b \ln(a)\). But at least I read \(\displaystyle \ln (a)^b\) as \(\displaystyle (\ln a)^b\). And for this we know \(\displaystyle (\ln a)^b \ne b \ln (a)\).

why would you read \(\displaystyle \ln (a)^b\) as \(\displaystyle (\ln a)^b\) this notation clearly is not the same thing?

 

[Theia]

why would you read \(\displaystyle \ln (a)^b\) as \(\displaystyle (\ln a)^b\) this notation clearly is not the same thing?

Because it looks like the whole argument of the logarithm is in parenthesis, and the exponent is not there. And so the possibility for ambiguity arises...

 

[MarkFL]

Because it looks like the whole argument of the logarithm is in parenthesis, and the exponent is not there. And so the possibility for ambiguity arises...

Yes, I agree completely and I would certainly prefer the notation:

\(\displaystyle \log_a\left(b^c\right)\)

This is clear where the exponent belongs. :D

 

[karush]

Yes, I agree completely and I would certainly prefer the notation:

\(\displaystyle \log_a\left(b^c\right)\)

This is clear where the exponent belongs. :D

it might be preferable but I see both all the time it not that confusing...

$\log(abc)^2$ or $log(a^2 b^2 c^2)$ but this $(log abc)^2$ is obviously different...

bone for the day award;)

 

[MarkFL]

it might be preferable but I see both all the time it not that confusing...

$\log(abc)^2$ or $log(a^2 b^2 c^2)$ but this $(log abc)^2$ is obviously different...

bone for the day award;)

Personally, for clarity, I would write:

\(\displaystyle \log\left((abc)^2\right)\)

For any function, I want anything and everything regarding the argument(s) inside brackets.