Find the general solution to the differential equation $\frac{dy}{dx} = \frac{2\sqrt{1+e^y}}{ \sec(x)} \cdot e^{\sin(x)-y}$

Question

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!

Answer 1

$\frac{1}{2}\frac{d(e^{y})}{\sqrt{1+e^{y}}} = e^{sin(x)}d(sin(x)))$
$\sqrt{1+e^{y}} = e^{sin(x)} + C$

Answer 2

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?

Answer 3

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!