variable saparable(separation of variable)

1.) (2x+y)dy + (2x+y+6)dx=0

2.)(5t+1)t ds + (25t-1)sdt=0

I don't the solution of this problem I think it a "repeated expression". Can someone help me on this? thanks a lot. :)

I noticed your equations when you posted them several days ago. I do not see how they can be solved by separation of variables. There are too many terms that involve both x and y.

Both problems are too difficult for me to find a closed-form solution.

2) (5t+1)t ds + (25t-1)sdt=0

Divide by (5t+1)ts :

ds/s + (25t-1)/[(5t+1)t]dt=0

1) (2x+y)dy + (2x+y+6)dx=0

Change of variables:

u = 2x+y

v = x

Then

y = u - 2 v

x = v

dy = du - 2 dv

dx = dv

1) becomes:

u (du - 2 dv) + (u+6)dv = 0 ------>

u du + (6 - u)dv = 0

Divide by (6-u):

u/(6-u) du + dv = 0

To solve these differential equations using separation of variables, we need to rearrange each equation so that all the terms involving the independent variable (x or t) are on one side, and all the terms involving the dependent variable (y or s) are on the other side.

Let's solve each equation step by step:

1) (2x+y)dy + (2x+y+6)dx = 0

First, let's rearrange the equation to separate the variables:
(2x+y)dy = -(2x+y+6)dx

To separate the variables, divide both sides by (2x+y+6):
(dy)/(2x+y) = -dx/(2x+y+6)

Now, let's integrate both sides with respect to their respective variables. On the left side, the integral is taken with respect to y, and on the right side, the integral is taken with respect to x.

∫(dy)/(2x+y) = ∫-dx/(2x+y+6)

Integrating the left side requires a substitution. Let u = 2x+y, then du = 2dx. Substituting back, we have:

(1/2) ∫(dy)/u = -∫(dx)/(u+6)

(1/2) ln|u| = -ln|u+6| + C1

Where C1 is the constant of integration. Simplifying further:

ln|u|^(1/2) = ln|u+6|^(-1) + C1

Take the exponential of both sides:

|u|^(1/2) = |u+6|^(-1) * e^(C1)

Squaring both sides to remove the absolute values:

u = 1/(u+6) * e^(2*C1)

Substituting back u = 2x+y:

2x+y = 1/(2x+y+6) * e^(2*C1)

2)(5t+1)t ds + (25t-1)sdt = 0

First, let's rearrange the equation to separate the variables:
(5t+1)t ds = -(25t-1)sdt

To separate the variables, divide both sides by (25t-1):
(t ds)/(25t-1) = -s dt/(5t+1)

Now, let's integrate both sides with respect to their respective variables. On the left side, the integral is taken with respect to s, and on the right side, the integral is taken with respect to t.

∫(t ds)/(25t-1) = ∫-s dt/(5t+1)

Integrating the left side requires a substitution. Let u = 25t-1, then du = 25dt. Substituting back, we have:

(1/25) ∫(t ds)/u = -∫(s dt)/(5t+1)

(1/25) ln|u| = -ln|5t+1| + C2

Where C2 is the constant of integration. Simplifying further:

ln|u|^(1/25) = ln|5t+1|^(-1) + C2

Take the exponential of both sides:

|u|^(1/25) = |5t+1|^(-1) * e^(C2)

Squaring both sides to remove the absolute values:

u = 1/(5t+1) * e^(25*C2)

Substituting back u = 25t-1:

25t-1 = 1/(5t+1) * e^(25*C2)

These are the solutions to the given differential equations using separation of variables.