Question

(4x+6y+5)dy=(3y+2x+4)dx

Answer 1

dy/dx = (3y + 2x + 4)/(4x + 6y + 5)

Let v = 2x + 3y, then dv/dx = 2 + 3*dy/dx

i.e. dy/dx = (1/3)(dv/dx - 2)

Substituting for v and dy/dx into the original DE,

(1/3)(dv/dx - 2) = (v + 4)/(2v + 5)

dv/dx - 2 = (3v + 12)/(2v + 5)

dv/dx = (3v + 12)/(2v + 5) + (4v + 10)/(2v + 5) = (7v + 22)/(2v + 5)

dv/dx = (7v + 22)/(2v + 5)

rearranging,

int (2v + 5)/(7v + 22) dv = int 1 dx

(2/7)*int (14v + 35)/(14v + 44) dv = int 1 dx

int 1 - 9/(14v + 44) dv = int (7/2) dx

integrating,

v - (9/14).ln(14v + 44) = (7x/2) (since seeking particular solution only, then ignore constant of integration)

substituting back for v = 2x + 3y

2x + 3y - (9/14).ln(14(2x + 3y) + 44) = 7x/2

28x + 42y - 9.ln(28x + 42y + 44) = 49x

42y - 21x = 9.ln(28x + 42y + 44)

Answer: 14y - 7x = 3.ln(28x + 42y + 44)

Answer 2

Step-by-step explanation:

(2x+y+1)dx=(4x+2y-1)dy