Find the general solution to the differential equation $\frac{dy}{dx} = \frac{2\sqrt{1+e^y}}{ \sec(x)} \cdot e^{\sin(x)-y}$ This is for Calculus 2.
Finding the General solution for:
$$\frac{dy}{dx} = \frac{2\sqrt{1+e^y}}{ \sec(x)} \cdot e^{\sin(x)-y}$$
Hello everyone,
I am not quite sure how to start solving this equation.
If someone could help me set up the equation by having the y-values on the the left. That is all I need.
Thank you!
 A: $\frac{1}{2}\frac{d(e^{y})}{\sqrt{1+e^{y}}} = e^{sin(x)}d(sin(x)))$
$\sqrt{1+e^{y}} = e^{sin(x)} + C$
A: Hopefully you can see that
$${2\sqrt{1+e^y}\over\sec x}e^{\sin(x)-y}=2(e^{-y}\sqrt{1+e^y})(e^{\sin x}\cos x)$$
Now move the stuff with $y$'s to the left hand side with the $dy$, and the $dx$ over to the right hand side. Is that enough to get you started?
A: To further supplement the others' answers you can always check for separability based on this equation:
$$\frac{dy}{dx}=f(x)\cdot g(y)$$
where $f$ and $g$ are functions of only $x$  and $y$ respectively.
Equation of this form can always be solved by Seperation of Variables by the rules of basic algebra (and a pinch of abuse of notation):
$$\dfrac{dy}{g(y)}=f(x)dx$$
Now all you need to do is integrate with respect to to $y$ and $x$.
$$\int\dfrac{dy}{g(y)}=?\quad\text{ or }\quad\int f(x)dx=?$$
Of course there is a chance that your functions don't have an elementary integral and you have to 'stop' at this step and resort to different toolery.
Eg. $$\int {\displaystyle {\sqrt {1-x^{4}}}}dx=???$$
If you manage to integrate:
$$G(y)=F(x)+C$$
where $G(y)$ is the primitive function of $1/g(y)$ and $F(x)$ is the primitive function of $f(x)$.
This is called the implicit solution, because $y$ is still kind of hidden, now you just need to do some algebra to get back $y$:
$$y=h((F(x)+C),(\text{stuff from the other side}))$$
Which is the explicit solution, this doesn't always exist either.
Eg. $$y+\cos(y)e^y+\ln(\cos(y))=5xe^x$$
Imagine if your problem told you $y(x)=w'(x)=\frac{dw}{dx}$ and it wanted you to get the answer in $w$'s:
$$\frac{dy}{dx}=f(x)\cdot g(y)$$
becomes
$$\frac{d(w')}{dx}=f(x)\cdot g(w')$$
$$\dfrac{d(w')}{g(w')}=f(x)dx$$
After integration you are left with:
$$G(w')=F(x)+C$$
which after solving for $w'$ you get
$$w'=\mathfrak{F}(x)$$
where $$\mathfrak{F}=\mathfrak{F}((F(x)+C),(\text{stuff from the other side}))$$
But we know $$w'=dw/dx$$
which is:
$$\frac{dw}{dx}=\mathfrak{F}(x)$$
whiches solutions we already know!
\begin{align}
\int dw & =\int \mathfrak{F}(x)\;dx\\
w & =\phi(x)+D
\end{align}
Thus we can belive that
$$\frac{d^2x}{dx^2}=f(x)g(y)$$
always can be separated.
In fact this property is true for such ODEs of order $n$ where you have to integrate $n$ times.
Beware: you can't just integrate however many times and just add together all the constants! The constants may pick up factors of $x$ or become the internatl parts of complex functions!
