Dy Dx Y 2 9
Dy Dx Y 2 9
Using the subtraction equation, solve for dy / dx = y
They are separable, that is, separate:
dy / (yò 9) = dx
Then integration. To your right is x + C. The left side needs some work. The denominator is divided by (y 3) (y + 3), so use a fraction.
A / (y 3) + B / (y + 3) = 1 / [(y 3) (y + 3)]
This is necessary for it to be true.
A (y + 3) + B (y 3) = 1
(A + B) y + 3 (A B) = 1 ... then set the number to the power equal to y
A + B = 0
3 (A B) = 1
Since B = A, then
3 (A (A)) = 1
A = 1/6
B = A = 1/6
dy / (y² 9) = (1/6) [dy / (y3) dy / (y + 3)]
«Dy / (y² 9) = (1/6) [ln | And 3 | In | And +3 | ] = (1/6) ln (| y 3 | / | y + 3 |]
Since we need a solution with (0,0), where y3 <0 and y + 3> 0, the absolute value can be omitted to obtain:
(1/6) ln [(3 years) / (y + 3)] = x + C
GACK! This is the left inverted stem. I think the hyperbolic alternative will work too. Now only »Ã‚ t = (1/2) ln ((1 + t) / (1 t)) = (1/2) ln ((1 t) / (1 + trt)),. And we make it easy:
(1/6) ln [(1 y / 3) / (1 + y / 3)] = x + C ... Divide the digit and denominator by 3 in ln []
1/2 ln [(1 and / 3) / (1 + y / 3)] = 3x + C ... Multiply both sides by 3, but 3C is still the arterial of C is constant.
tanha »Ã‚¹ (y / 3) = 3x + C
Such as tanh (0) = 0, C = 0. It can be solved as a function of x to get y. Take the earth from both sides:
y / 3 = tanh (3x)
y = 3 years (3x) = 3 years (3x)
Give it a try y = 3 [sec² (3x) * (3)] = 9 sec² (3x)
= 9 (1 tanh² 3x) = 9 (tanh² 3x 1) = 9 tanh (3x) 9 = y² 9
It worked.
Because:
dy / dx = y² 9 y (0) = 0
Equations can be separated:
{1 / (y² 9) dy = dx y (0) = 0
Connect both sides:
X "{1 / (y² 9) dy = ˆÂ" dx y (0) = 0
Since y² 9 is counted in (y + 3) (y 3), we use the extension of the partial fraction to make the left side:
A / (y + 3) + B / (y 3) = 1 / {(y + 3) (y 3)
A (y 3) + B (y + 3) = 1
3A + 3B = 1
Ay + straight line = 0
B = A
3A 3A = 1
6A = 1
A = 1/6
B = 1/6
(1/6) "{1 / (y + 3)} dy + (1/6) ˆÂ" {1 / (y 3)} dy = ˆ "dx y (0) = 0
(1/6) in | And +3 | + (1/6) inch. | And 3 | = x + Cy (0) = 0
In | (y 3) / (y + 3) | = 6x + C and (0) = 0
(y 3) / (y + 3) = Ce (6x) y (0) = 0
(y + 3 3 3) / (y + 3) = Ce (6x) y (0) = 0
1 6 / (y + 3) = Ce (6x) y (0) = 0
6 / (y + 3) = Ce (6x) 1 y (0) = 0
(y + 3) / 6 = 1 / {Ce (6x) 1} y (0) = 0
y + 3 = 6 / {C and (6x) 1} y (0) = 0
y = 6 / {C and (6x) 1} 3 y (0) = 0
Diagnosis of threshold condition:
0 = 6 / {C1} 3
3 = 6 / {C1
3C 3 = 6
3C = 6
C = 1
Replace 1 with C:
y = 6 / {e (6x) + 1} 3
dy / dx = y 29, y (0) = 0
First, we need to separate the variables. Divide by (y 29) and multiply by dx:
dy / (y 29) = dx
Now connect the two sides.
Integration of dy / (y 29) Triangles (which I will check in your o):
1/3 * Arkanh (J / 3)
Integral on the X-Side is easy, remember the permanent and we get:
1/3 * arctanh (y / 3) = x + C
Now we solve for y. Multiply by 3:
arctanh (y / 3) = 3x + C
Take land on each side:
y / 3 = tanh (3x + C)
Multiply by 3:
y = 3tanh (3x + C)
To find the relevant solution, we use y (0) = 0 and add the two to it:
0 = 3tanh (3 (0) + C)
Divide by 3 and multiply / increase the hyperbolic ent:
0 = Tan (C)
Using reverse activation for hyperbolic ent, we find that C is 0, so the final solution is:
y = 3tanh (3x)
So you are trying to find a specific solution with 0.0 point of this function and
Separate the first variables: 1 / y 2 x dy = 9 dx
Then take the antiderivative: 1 / y = 9x + C (remember there is always a constant)
Enter the given values to find the value of C: 1/0 = 0 + C C = 0
Solve for y and get the equation ure: 1 / y = 9x + 0
y = 1/9 x
YAYYAYAYAYAYA asks what helped (you should always follow these four steps when trying to resolve your differences).
Dy Dx Y 2 9
Dy Dx Y 2 9
Find a general solution by separating and integrating the variables:
dy / dx = y² 9
dy / dx = (9 y²)
dy / (9 y²) = 1 dx
1 "1 / (9 years) dy =" 1 dx
tanha ¹ (y / 3) / 3 = x + C
tanha ¹ (y / 3) / 3 = C x
tanha  (y / 3) = C 3x
y / 3 = tanh (C 3x)
y = 3tanh (C 3x)
Find specific solutions by solving permanent:
If x = 0, y = 0
3tanhC = 0
tanhC = 0
C = 0
y = 3tanh (3x)
y = 3tanh (3x)
Use of subtraction equations: it doesn't make sense!
Just solve the problem.
Merge once and get y = (1/3) tanh 1 (x / 3)
Check if this is the correct answer.
dy / dx = y 29
=> dy / (y 29) = dx
For integration,
(1/6) ln l (y3) / (y + 3) l = x + c
y (0) = 0
=> c = 0
=> (1/6) ln l (y 3) / y + 3) l = x
=> [(y 3) / (y + 3)] = e (6x)
=> [(y + 3) (y 3)] / [(y + 3) + (y 3)] = [1 and (6x)] / [and ( x) + 1]
=> y = 3 [1 and (6x)] / [e (6x) + 1].
The s-of-4-minus balance of principal 67743.89 is part of the free-circulation condition. Produces an average square content of 5.8 on a regular basis with a degree of light disturbance !!! ... You told him that Raven called !! The fact that !!
