# [SOLVED]242.7x.13 Solve the differential equation

#### karush

##### Well-known member
$\tiny{242.7x.13}$
$\textsf{Solve the differential equation}$
\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

Staff member
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.

#### karush

##### Well-known member
$\tiny{242.7x.13}$
$\textsf{Solve the differential equation}$
\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

Staff member
$\tiny{242.7x.13}$
$\textsf{Solve the differential equation}$
\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.

#### karush

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

Last edited:

#### MarkFL

Staff member
\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

##### Well-known member
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
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
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

Staff member
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.

#### karush

##### Well-known member
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.
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

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