Welcome to our community

Be a part of something great, join today!

[SOLVED] 242.7x.13 Solve the differential equation

karush

Well-known member
Jan 31, 2012
2,928
$\tiny{242.7x.13}$
$\textsf{Solve the differential equation}$
\begin{align*}\displaystyle
\frac{dy}{dx}&=7x^6e^{-y}\\
dy&=7x^6e^{-y} \, dx \\
e^y dy&=7x^6 dx
\end{align*}
$\textit{just seeing if going the right direction on this...}$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Jan 31, 2012
2,928
$\tiny{242.7x.13}$
$\textsf{Solve the differential equation}$
\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

Administrator
Staff member
Feb 24, 2012
13,775
$\tiny{242.7x.13}$
$\textsf{Solve the differential equation}$
\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

Well-known member
Jan 31, 2012
2,928
\begin{align*}\displaystyle
ln (e)^y&=ln\left({x}\right)^7 \\
y&=7ln(x)+C
\end{align*}
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\begin{align*}\displaystyle
ln (e)^y&=ln\left({x}\right)^7 \\
y&=7ln(x)+C
\end{align*}
$\text{easy to forget our C friend...}:cool:$
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

Well-known member
Mar 30, 2016
115
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

Well-known member
Jan 31, 2012
2,928
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

Well-known member
Mar 30, 2016
115
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Jan 31, 2012
2,928
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

Administrator
Staff member
Feb 24, 2012
13,775
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.