1. ## square root equations

A.
$xdy-ydx=\sqrt{x^{2}+y^{2}}dx$
B.
$\sqrt{y^{2}+1}dx=xydy$

how to solve these

2. Originally Posted by transgalactic
B.
$\sqrt{y^{2}+1}dx=xydy$

how to solve these
$\sqrt{y^{2}+1}~dx=xy~dy$

$\frac{\sqrt{y^{2}+1}}{x}~dx=y~dy$

$\frac{dx}{x}=\frac{y}{\sqrt{y^{2}+1}}~dy$

$\int \frac{dx}{x}=\int \frac{y}{\sqrt{y^{2}+1}}~dy$

Use a substitution $u = y^{2}+1$ on the RHS.

3. Originally Posted by transgalactic
A.
$xdy-ydx=\sqrt{x^{2}+y^{2}}$
B.
$\sqrt{y^{2}+1}dx=xydy$

how to solve these
I believe that you've stated A incorrectly. Please check.

4. you are right its
$xdy-ydx=\sqrt{x^{2}+y^{2}}dx$

how to solve it?

5. This is homogeneous. Try letting $y = xu$.

6. how did you know its homogeneous and you need to put
y/x=u

??
dy=xdu
$x^2du-xudx=\sqrt{x^2+x^2u^2}dx$
what now?
i cant put on one side dy on the other dx
but its not looking like something solveable
?

7. No $y = xu$ and $dy = x du + u dx$. Substitute these and simplify. You should get something separable.

8. i did this substitution
i got
$x^2 du=\sqrt{x^2+x^2u^2}dx$
how to solve it

9. still i got a non saparable equation

??

10. Originally Posted by transgalactic
i did this substitution
i got
$x^2 du=\sqrt{x^2+x^2u^2}dx$
how to solve it
You need to factor

$x^2 du=\sqrt{x^2+x^2u^2}dx$

becomes

$x^2 du=x\sqrt{1+u^2}dx$

Now separate.

11. Originally Posted by transgalactic
i did this substitution
i got
$x^2 du=\sqrt{x^2+x^2u^2}dx$
how to solve it
$x^2 du= \sqrt{x^2(1+ u^2)}dx$
That certainly is separable.

12. thanks i see that now