Differential Equation

Question

What is the particular solution to the differential equation dy/dx=18(2+y)^2​/(1−3x)^2 with the initial condition y(1)=−4?

Answer 1

separation:

dy/(y+2)^2 = [18 dx]/(1-3x)^2

integrates both sides

U = y+2 ---> dU = dy

V = 1-3x --> dV = -3 dx

dU/ U^2 = -6 dV/ V^2

U^(-2) dU = -6 V^(-2) dV

-U^(-1) = 6 V^(-1) + c

-1/(y+2) = 6/(1-3x) + c

Multiplies by (y+2)(1-3x):

3x-1 = 6(y+2)+ c(1-3x)(y+2)

3x-1 = [6 + c(1-3x)] (y+2)

(3x-1)/ [ 6 + c(1-3x)] = y+2

(3x-1)/[6 + c(1-3x)] - 2 = y

(3x-1)/[6 + c - 3cx] - 2 = y

Applying the intitial condition:

2/[ 6 +c - 3c] - 2 = -4

2/[6 - 2c] - 2 = -4

1/[3 - c] - 2 = -4

1/[3-c]= -2

1 = 2(c-3)

1 = 2c - 6

7 = 2c

c = 7/2

Then solution is

y = [(3x-1)/(19/2 - (21/2)x)] - 2

Answer 2

We separate variables and integrate: ∫dy/(2 + y)² = 18∫dx/(1 - 3x)²; - 1/(2 + y) = 6/(1 - 3x) + C;

To find C we use initial conditions: - 1/(2 - 4) = 6/(1 - 3) + C; C = 1/2 + 3 = 7/2;

1/(2 + y) = 6/(3x - 1) - 7/2.

If you like solve this for y

Answer 3

Variables are separable:

(-1/18)([1/(2+y)]=(1/3)[1/(1-3x)]+C

I will let you do the initial condition part to evaluate C.