(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)...

42

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

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

This is where you use integrating factor right?

General method for a LODE (linear ordinary differential equation) :

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

First consider the homogenuous LODE (HLODE)

(HLODE) : y' + a(x).y = 0

Find a solution for (HLODE) of the form : y₀(x) = K. exp(-A(x))
where A is an anti-derivative of a.

Example with (1) : x.y' = x² + 3y
(1) : y' = x + 3 y/x
(LODE) : y' - (3/x).y = x
(HLODE) : y' - (3/x).y = 0

a(x) = - 3/x
A(x) = - 3 . log x

y₀ = K. exp (3. log x) = K. x^3

Now try to find a specific solution for (LODE) by considering that the constant K is no longer constant : K -> K(x)

y₁ (x) = K(x). x^3 is a solution of (LODE) if and only if : y₁ ' = x + 3 y₁/x
with :
y₁ '(x) = K'(x). x^3 + 3 K(x). x²

In turn :
K'(x). x^3 + 3 K(x). x² = x + 3.K(x). x^3 / x
iff
K'(x).x^3 = x
iff
K'(x) = 1/x²
iff
K(x) = - 1/x

That should give you a specific solution : y₁ (x) = (-1/x). x^3 = - x^2

(at that point, you can check that y₁ works)

Conclusion : the general form of a solution for (LODE) is y = y₀ + y₁ :
y(x) = K. x^3 - x^2
where K is a real constant.

i just used integating factor for 1 and it wroked lol... I got y = - x^2 + x^3 *c OR y/x^3 = -1/x + c

#2 im stuck on,

its not exact, separable, homogenous

This post was edited by FamilyGuyViewer on Sep 5 2016 02:03pm

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

...

This post was edited by feanur on Sep 5 2016 02:30pm

hold on, tryna do 9 right now.

missingg 2 ,7 9

for 7.

i got 2(x/y) + x^2/2 + 2 ln y = constant

fck it, i just copied the end result for #2, annoying ass integrals.