What is the general solution of the differential equation? # (4x+3y+15)dx + (2x + y +7)dy = 0 #
1 Answer
# 17/34 ln abs(((y+1)/(x+3))^2-(y+1)/(x+3)-4) - 5sqrt(17)/34 ln abs(2(y+1)/(x+3)+sqrt(17)-1) + 5sqrt(17)/34 ln abs(2 (y+1)/(x+3)-sqrt(17)-1) = ln abs(u) + C #
Explanation:
We have:
# (4x+3y+15)dx + (2x + y +7)dy = 0 #
Which we can write as:
# dy/dx = - (4x + 3y + 15)/(2x +y +7) # ..... [A]
Our standard toolkit for DE's cannot be used. However we can perform a transformation to remove the constants from the linear numerator and denominator.
Consider the simultaneous equations
# { ( 4x + 3y + 15=0 ), (2x + y +7=0) :} => { ( x=-3 ), (y=-1) :} #
As a result we perform two linear transformations:
Let
# { (u=x+3 ), (v=y+1) :} => { ( x=u-3 ), (y=v-1) :} #
And if we substitute into the DE [A] we get
# (dv)/(du) = - (4(u-3) + 3(v-1) + 15)/(2(u-3) +(v-1) +7) #
# \ \ \ \ \ \ = - (4u-12 + 3v-3 + 15)/(2u-6 +v-1 +7) #
# \ \ \ \ \ \ = - (4u+ 3v)/(2u+v) # ..... [B]
This is now in a form that we can handle using a substitution of the form
# (dv)/(du) = (w)(d/(du)u) + (d/(du)w)(u) = w + u(dw)/(du) #
Using this substitution into our modified DE [B] we get:
# \ \ \ \ \ w + u(dw)/(du) = - (4u+ 3wu)/(2u+wu) #
# :. w + u(dw)/(du) = - (4+3w)/(2+w) #
# :. u(dw)/(du) = - (4+3w)/(2+w) -w #
# :. u(dw)/(du) = - ( (4+3w) - w(2+w) ) / (2+w) #
This is now a separable DE, so we can rearrange and separate the variables to get:
# int \ (2+w)/( w^2-w-4 ) \ dw = int \ 1/u \ du #
The LHS is a standard integral, and the RHGS integral can be split into partial fractions (omitted) and then it is readily integrable, and we find that:
# 17/34 ln abs(w^2-w-4) - 5sqrt(17)/34 ln abs(2w+sqrt(17)-1) +5sqrt(17)/34 ln abs(2 w-sqrt(17)-1) = ln abs(u) + C #
Then restoring the earlier substitution we have:
# w=v/u = (y+1)/(x+3) #
Thus:
# 17/34 ln abs(((y+1)/(x+3))^2-(y+1)/(x+3)-4) - 5sqrt(17)/34 ln abs(2(y+1)/(x+3)+sqrt(17)-1) + 5sqrt(17)/34 ln abs(2 (y+1)/(x+3)-sqrt(17)-1) = ln abs(u) + C #
This can be further simplified if desired.
