Quote (FaceDeath @ Sep 6 2016 03:57pm)

pm me
i can do this

Quote (feanur @ Sep 5 2016 09:47am)

(1) to (4), and (9) are of the form :

y' + a(x).y = b(x)

Linear ODE of order 1.
The method should be known (not all of them are easy to solve, however).

(5) is separable : 1/x = y' y² / (1-y^3)

(8) requires a change of variable : let t=y/x
y = tx, dy = tdx + xdt
dy/dx = t + x dt/dx

(8) gives dy/dx = 1/cos t + t
t + x dt/dx = 1/cos t + t
x dt/dx = 1/cos t
which is now separable :

t' cos t = 1/x
sin t = ln x + K
t = Arcsin (ln x + K)
y(x) = x . Arcsin ( ln x + K )


What I have for (6) :
(sin y - y sin xy ) dx + ( x cos y - x sin xy ) dy = 0
dx sin y + x dy cos y = sin xy ( y dx + x dy )
d ( x sin y) = sin xy d(xy)

Let t = xy :
d ( x sin y) = sin t dt = d ( - cos t)
x sin y = - cos t + K = - cos (xy) + K

which is an implicit solution :
x sin y + cos (xy) = K

I must say I am stuck on (7)...

Quote (feanur @ Sep 5 2016 09:29pm)

Just thought a second time at (7) : y(1+xy)dx + (2y-x) dy = 0

Expand :
ydx + xy²dx + 2ydy - xdy = 0

Divide by y² :

(ydx - xdy)/y² + xdx + 2dy/y = 0

d(x/y) + (1/2) d(x²) + 2 d(log y) = 0

x/y + x²/2 + 2 log y = Constant

This is an implicit form, I guess you are not expected to go further.


For (2) : 3xy' - y = log x + 1 (with initial condition)

(LODE) : y' - (1/3x) y = (log x + 1) / (3x)
(HLODE) : y' - (1/3x) y = 0

y₀(x) = K . exp ( (1/3) log x ) = K . x^(1/3)

y₁(x) = K(x) . x^(1/3)
y₁ ' (x) = K ' (x) . x^(1/3) + (1/3) K(x) x^(-2/3)

y₁ is a solution of (LODE) iff :
K ' (x) . x^(1/3) + (1/3) K(x) x^(-2/3) - (1/3x) K(x) . x^(1/3) = (log x + 1) / (3x)
K ' (x) . x^(1/3) = (log x + 1) / (3x)
K ' (x) = (log x + 1) / (3 x^(4/3) )

Now to find K (x) :
K (x) = ∫ (log x + 1) dx / (3 x^(4/3) ) = (1/3) ∫ log x dx / x^(4/3) + (1/3) ∫ dx / x^(4/3)

The second ∫ is easy.
For the first, try an integration by parts :
∫ log x dx / x^(4/3) = [ -3 x^(-1/3) log x ] - ∫ (-3) x^(-1/3) (1/x) dx

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