god i feel dirty for doing this but i have a total mental block on two problems so i just need a point in the right direction.

 

number 1 is:

 

Find a solution of the differential equation

 

Dy/Dx=(2x^2+y^2)/(2xy) (x>0 , y>0)

 

which satisfies y=1 and x=1

 

and number 2 is:

 

(1+x)Dy/Dx + 4y = 1/(x+1) (x (cannot equal sign) -1)

First one is possibly seperation of variables, but the easiest option is probably using an integration factor. I *think* you should be able to get it into the correct form. I have some quick notes on the method for solving first order differntial equations on my other computer I'll pull them out and post them later. Thinking about this one there is a third method called the exact method that might work, I shall check my notes later and get back to you...

 

This second one is most certainly an integrating factor ODE, it's just getting it in the correct form P(x)dy+Q(x)ydx = 0 and then solving a set of simple equations.

i've managed to get the second one (it was an integrating factor) but i can't seem to get the first one in the right form

Is the first one really dy/dx = (2x² + y²)/2xy?

 

When it is dy/dx = (2x² - y²)/2xy, then it is easy. Otherwise it is a hard problem.

its really + and its nice to see that i'm not struggling with an easy problem but a hard one.

Yeah it would be exact if it was a -ve, looks like you'll have to solve the characteristic equation!

 

Good luck

Well, I see still no solution. I'll give a few hints, but you'll have to derive the solution yourself.

 

dy/dx = (2x² + y²) / 2xy

 

Reorganize and integrate in y:

 

2xy dy = 2x²dx + y²dx

 

xd(y²) = 2x²dx + y²dx

 

Now, substitute y² = z

 

xdz = 2x²dx + zdx

 

Now, with some rearranging, and dividing left and right by x, you have a linear equation in z, with non-constant coefficients and non-homogeneous:

 

dz/dx - (1/x)z = 2x

 

This one is easy to solve (I assume you know how to solve first order linear differential equations, even if they have non-constant coefficients and are non-homogeneous).

 

Try to continue from this point. I am willing to help you further with this, but first try yourself from this point. It still is not really easy, but I think you can do it from this point. Indeed, the original problem was not easy.

thanks guys i got it done now. i owe you all one.

Is your answer to the first one y = √(-x + 2x²) ?

 

The equation has solutions of the form y = √(Ax + 2x²), with any A > 0, but when you need the trajectory of the solution to go through (1,1), then A = -1.

 

Btw, what is meant with the term "integration factor" in this context? I learned solving this kind of equations more than 20 years ago, but at that time, this term was never used.