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.
