Question

http://s783.photobucket.com/albums/yy112/Chad_25_2009/?action=view¤t=DE.jpg

Answer 1

& No. 5 Pls...

Answer 2

I am going to solve #4 \((x+y)dx+(x-y)dy=0\). Let N(x,y)=x+y and M(x,y)=x-y. Then we have \(N_y=1\) and \(M_x=1\), so it's exact and there exists a function f(x,y) such that \(f_x(x,y)=N(x,y) ... (1)\) and \(f_y(x,y)=M(x,y) ... (2)\). By integrating (1) with respect to x we get:
\(f(x,y)={1 \over 2}x^2+yx+g(y) \) \((*)\)
Now, differentiate (*) with respect to y and get:
\[f_y=x+g'(y)=x-y \implies g'(y)=-y \implies g(y)=-{1 \over 2}y^2\]
Substitute for g(y) in (*) and we get:
\[f(x,y)={1 \over 2}x^2+yx-{1 \over 2}y^2\]. Hence the solution of the given differential equation is \(\frac{1}{2}x^2+yx-\frac{1}{2}y^2=c\).

Answer 3

Thank you so much & No. 5 Pls...

Answer 4

Are you sure about No. 5?. It's not a differential equation.

Answer 5

ok if it is not could you Obtain the Solution.?

Answer 6

Solution of what? You have to be more specific :)

Answer 7

if you have solution for that pls

Answer 8

If you're looking for solutions of differential equations, then that is already a solution. When we talk about differential equations, we talk about equations that contain a derivative or more, and then we try to solve it and find the original function. In this case, we don't have any derivatives, so I can't do anything with it.

Answer 9

ohh really in the No. 5 Question written in the photo :(

Answer 10

You may want to check the question with one of your classmates.

Answer 11

No i'm the only one

Answer 12

Then ask your professor :)

Answer 13

and i need that 2morrow is the submission

Answer 14

pls HELP ME IF YOU CAN