Question

find the general solution of equation
xdy-( sqrt(y^2+x^2)+y)dx=0

Answer 1

WolframAlpha has a well-explained solution:

Answer 2

thanks alot but i don't understand how do you get(sinh) in the answer;

Answer 3

xdy -ydx = sqrt(x^2+y^2)dy
-(-xdy +ydx)/y^2 = sqrt(x^2+y^2) /y^2 dy
or -d(x/y)= sqrt(1+(x/y)^2) /y dy
or -d(x/y)/ sqrt(1+(x/y)^2) =1/y dy
or ln c -ln(x/y +sqrt(1+(x/y)^2)) =ln y
and so on...

Answer 4

Read up until the step\[\int\limits \frac{ v'(x) dx}{ \sqrt{(v(x))^2+1} }=\int\limits \frac{ 1 }{ x }dx\]Now \[(\arcsinh(x))' =\frac{ 1 }{ \sqrt{x^2+1} }\]So with the Chain Rule (in reverse) the integrals become\[\arcsinh(v(x))=\ln|x| + c\]And\[v(x)=\sinh(\ln|x| +c)\]But: xv(x)=y(x), and this leads to:\[y(x)=x \sinh(\ln|x|+c)\]

Answer 5

thanks; I understood

Answer 6

I was trying to solving it
cheek this answer please;
let x=vy,dx=vdy+ydv
vydy-(sqrt(v^2 y^2)+y)(vdy+ydv)=0
-vy sqrt(v^+1)dy-y^2(sqrt(v^2+1)+1)dv=0
-y(vsqrt(v^2+1)dv+y(sqrt(v^2+1)dv))dv=0
let y not equal zero
by intergration
-ln|y|=ln|v|-arc csch(v)+c

Answer 7

I think you've forgot to put y² in the second line. It should be:

vydy-(sqrt(v²y²+y²)+y)(vdy+ydv)=0

Answer 8

Then:

vydy-(ysqrt(v²y²+1)+y)(vdy+ydv)=0

Answer 9

oh my god, that's right
Sorry for your trouble with me

Answer 10

thanks again

Answer 11

variation of parameters.

Answer 12

what do you mean?

Answer 13

Are you using variation of parameters to solve? If you don't know what I am talking about then just ignore me. Lol. I don't want to confuse you more.

Answer 14

I didn't use variation of parameter to solve this equation
thanks for your time