Differential Equation?

Question

Solve this differential equation

dy/dx = (x^2 + y^2) / 2xy

There are some algebra mistakes in Pascal's solution....

Answer 1

First, multiply both sides by 2y:

2y dy/dx = (x²+y²)/x

Now, make the substitution u=y², du/dx = 2y dy/dx:

du/dx = (x²+u)/x
du/dx - 1/x u = x

Dividing by x:

1/x du/dx - 1/x² u = 1

This is nothing more than the derivative of the following product:

d(1/x u)/dx = 1

Integrating both sides:

u/x = x+C

u = x² + Cx

Making the substitution:

y² = x² + Cx
y=±√(x² + Cx) (sign depends on the initial value)

And we are done.

Edit: Actually, the use of the plural is incorrect, as there was only one mistake in my solution. And it's not there anymore.

Answer 2

y = ±x*√(1-c/x) is the solution.
Proof:

dy/dx = (x^2 + y^2) / 2xy
dy/dx = x/(2y) + y/(2x)
Let t = y/x, then y = tx
and dy/dt = x*dt/dx + t

so x*dt/dx + t = 1/(2t) + t/2
x*dt/dx = 1/(2t) - t/2 = (1-t^2)/(2t)
(2t)/(1-t^2) dt = (1/x) dx
and the variables are separated.

∫ (2t)/(1-t^2) dt = ∫ (1/x) dx
∫ (2t)/(1-t^2) dt = lnx + c
Let u = (1-t^2); du = -2t dt
-∫ (1/u) du = lnx + C
-lnu = lnx + C
1/u = Cx
u = 1/(Cx) = c/x (c = 1/C)
1 - t^2 = c/x
1 - (y/x)^2 = c/x
(y/x)^2 = 1 - c/x
y/x = ±√(1 - c/x)
y = ±x√(1 - c/x)
{the sign depends on initial conditions}

You can always go back at this point and substitute this value into the original diff. eq. to see if it works.

y' = √(1 - c/x) + c/[2x√(1 - c/x)]
= [2x(1-c/x) + c] / [2x√(1 - c/x)]
= (2x-c) / [2x√(1 - c/x)]
and
(x^2 + y^2) / 2xy = (x^2 + x^2(1-c/x)) / [2x*x√(1 - c/x)]
= (2x^2 - cx) / [2x^2√(1 - c/x)]
= (2x-c) / [2x√(1 - c/x)]

Answer 3

the above one is a homogenous differential equation of 2nd order
for all such equations,put y = vx
differentating on both sides,
dy/dx = v + x dv/dx
but dy/dx = x^2 + y^2/2xy
= x^2 + v^2 * x^2/2x(vx)
=x^2(1+v^2)/2vx^2
=(1+v^2)/2v
implies v + x dv/dx = (1+v^2)/2v
x dv/dx = (1+v^2-2v^2)/2v
x dv/dx = (1-v^2)/2v
2v/(1-v^2) dv = dx/x
integrating on both sides,
integration of dx/x is logx + c
integration of 2v/(1-v^2)
let 1-v^2 = u
differentating both sides
-2v dv = du
integration of 2v/(1-v^2)dv is integration of
[ -du/u]
= -logu
= -log(1-v^2)
substituting the value of v,
= -log(1-y^2/x^2) = -log(x^2-y^2/x^2)
i.e. -log(x^2-y^2/x^2) = logx + logc
logx + log(x^2-y^2/x^2) = constant
logx[(x^2-y^2)/x^2] = -logc
since log ab = loga + logb
log[x^2-y^2]/x+logc =0
log{c[x^2-y^2]/x} = 0
c(x^2-y^2)/x = 1
x^2- y^2 = x/c(let 1/c=k some constant)
hence the solution of the differential equation is
x^2- y^2 = kx

Answer 4

dude, there's a formula for that type of Dif. Eq. look in your book.

Answer 5

dy/dx= x^2+y^2/2xy= (2x+2y)2xy-2y(x^2+y^2)/4x^2y^2 = 4x^2y+4y^2x-2yx^2-2y^3/4x^2y^2= 4y^2x+2yx^2-2y^3/4x^2y^2
try your self more ok
its just simplification