Question

Solve the separable differential equation:

9x-4y sqrt(x^2+1) dy/dx = 0

Subject to the initial condition: y(0)=3.

Answer 1

\[9x-4y \sqrt{x^2+1}\frac{ dy }{dx}=0\]

Answer 2

\(9x = 4y \sqrt{x^2+1}\frac{ dy }{dx}\)

\(\dfrac{9x}{ \sqrt{x^2+1} } = 4y \frac{ dy }{dx}\)

then do !

Answer 3

pattern on LHS is \(\dfrac{d}{dx} \sqrt {f(x)} = \dots\)

Answer 4

Thank you, but what do you mean "pattern on LHS"?

Answer 5

\[\implies \int \left ( \dfrac{9x}{ \sqrt{x^2+1} } = 4y \frac{ dy }{dx} \right) \color{red}{dx}\]

@SolomonZelman best guy/gal at explaining this, IMHO :-))

Answer 6

Yes, the equation is separated as follows (as IrishBoy123 showed).
\(\color{black}{\displaystyle \frac{9x}{\sqrt{x^2+1}}=4y \frac{ dy }{dx}}\)

then, you integrate both sides wrt x,
\(\color{black}{\displaystyle \color{red}{\int}\frac{9x}{\sqrt{x^2+1}}\color{red}{dx}=\color{red}{\int}4y \frac{ dy }{dx}}\color{red}{dx}\)

\(\color{black}{\displaystyle \int\frac{9x}{\sqrt{x^2+1}}dx=\int 4y~dy}\)

--------------------------------------------------------------
(This is why, btw, when you solve equations in the form
\(\color{black}{\displaystyle f(x)=g(y)~\frac{dy}{dx} }\) you integrate wrt x and wrt y on each side, as follows,

\(\color{black}{\displaystyle \int f(x)~dx=\int g(y)~dy}\)

then in general you would continue...
\(\color{black}{\displaystyle F(x)+C=G(y)}\) (capital lettered functions are antiderivaties)
The use your initial condition to solve for C ... etc.
---------------------------------------------------------------
The integral on the left side is by u-sub (you can do trig-sub as well, as a matter of a fact), and the right is obvious. The integral on the right is by power rule. Then, use your initial conditions to find C.

Answer 7

Nothing to add, other than I'm not the best source for DE material on this site ... hope this helps.

Answer 8

In terms of *Pattern*, via the power and chain rules....

\(\dfrac{d}{dx} (x^2 + 1)^{1/2} = \dfrac{1}{2} \cdot (x^2 + 1)^{\color{red}{-1/2}} \cdot 2x= \dfrac{x }{\sqrt{x^2 + 1}}\)


So

\(\int \dfrac{9x }{\sqrt{x^2 + 1}} ~ dx =9 \int \dfrac{x }{\sqrt{x^2 + 1}} ~ dx =9 \int \dfrac{d}{dx} (x^2 + 1)^{1/2} ~ dx\)

\( =9 (x^2 + 1)^{1/2} + C\)

It looks worse when it's written out longhand. But, in the head, it's way shorter.


You can achieve the same result in the manner of a **substitution**:

\(z = x^2 + 1, \color{blue}{dz} = 2x ~ \color{blue}{dx}\)


that often includes treating the components of the Liebnitz differential notation as components of a fraction, so...

\(9x-4y \sqrt{x^2+1} \dfrac{dy}{dx} = 0\)

\(\implies 9x=4y \sqrt{x^2+1} \color{red}{\dfrac{dy}{dx}}\)

\(\implies \color{red}{dx} ~ 9x=4y \sqrt{x^2+1} ~ \color{red}{dy}\)

I'm no expert but these ideas are waved around so much; and I wonder why.