Question

find the equation of the curve which satisfies the differential equation (1+y)dy+(1+x)dx=0 and passes through the origin.

Answer 1

Separate and integrate.
Move one of the terms to the other side and integrate both sides.

Answer 2

after integrateing what i do that's a problem

Answer 3

when you integrate, you'll get that +C at the end. Plug in x=0, y=0 to solve for C.

Answer 4

after intergrate i get y+\[\frac{ y2 }{ 2 }\]and x+\[\frac{ x2 }{ 2}\] is it write or worng

Answer 5

That's fine as long as that is all on one side of the equation. There is still the +C, so you can leave that on the other side of the equation, but you'll soon see that when you put in (0,0), C=0.

Answer 6

but ans. is x^2+y^2+2x+2y=0

Answer 7

Yeah, that'll work; after you have it set =0, you can multiply everything by 2 to clear the fractions.

Answer 8

i can't understand can you tell me in detail

Answer 9

I suppose. I'll show you my steps:
\[(1+y)dy+(1+x)dx=0\]
\[\int\limits (1+y)dy+ \int\limits (1+x)dx=0\]
\[y+\frac{y^2}{2}+x+\frac{x^2}{2}+C=0\]
\[x=0,y=0 \rightarrow 0+\frac{0^2}{2}+0+\frac{0^2}{2}+C=0 \rightarrow C=0\]

Answer 10

\[\rightarrow y+\frac{y^2}{2}+x+\frac{x^2}{2}=0\]
Multiply everything by 2, and the equation is:
\[2y+y^2+2x+x^2=0\]

Answer 11

You can verify by using implicit differentiation to find dy/dx of that equation, and you can also find dy/dx by rearranging the original differential equation, and you'll see that the two are equal.

Answer 12

thanks for help me

Answer 13

My pleasure.