Solve the following differential equation? #(2x+y+3)dy/dx=x+2y+1#
1 Answer
# y+x+4/3 = A(y-x-2)^3 #
Explanation:
We have:
# (2x+y+3)dy/dx = x+2y+1#
# :. dy/dx = (x+2y+1)/(2x+y+3) # ..... [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
# { ( x + 2y +1 =0 ), (2x +y + 3=0) :} => { ( x=-5/3 ), (y=1/3) :} #
As a result we perform two linear transformations:
Let
# { (u=x+5/3 ), (v=y-1/3) :} <=> { ( x=u-5/3 ), (y=v+1/3) :} => { ( (dx)/(du)=1 ), ((dy)/(dv)=1) :}#
And if we substitute into the DE [A] we get
# dy/dx = ((u-5/3)+2(v+1/3)+1)/(2(u-5/3)+(v+1/3)+3) #
# \ \ \ \ \ \ = (u+2v)/(2u+v) #
And utilising the chain rule we have:
# (dy)/(dx) = (dy)/(dv) (dv)/(du) (du)/(dx) => (dy)/(dx) = (dv)/(du) #
Thus we have a transformed equation
# (dv)/(du) = (u+2v)/(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) = (u+2wu)/(2u+wu) #
# :. w + u(dw)/(du) = (u+2wu)/(2u+wu) #
# :. u(dw)/(du) = (u+2wu)/(2u+wu) - w #
# :. u(dw)/(du) = ( (u+2wu) - w(2u+wu) ) / (2u+wu) #
# :. u(dw)/(du) = ( u+2wu - 2uw-w^2u ) / (2u+wu) #
# :. u(dw)/(du) = ( u(1-w^2) ) / (u(2+w)) #
# :. u(dw)/(du) = (1-w^2) / (2+w) #
This is now a separable DE, so we can rearrange and separate the variables to get:
# int \ (2+w)/(1-w^2) \ dw = int \ 1/u \ du #
# :. int \ (2)/(1-w^2)+(w)/(1-w^2) \ dw = int \ 1/u \ du #
# :. int \ (2)/((1+w)(1-w))+(w)/(1-w^2) \ dw = int \ 1/u \ du #
And utilising a Partial Fraction decomposition:
# int \ 1/(w+1)-1/(w-1)+(w)/(1-w^2) \ dw = int \ 1/u \ du #
Which is now readily integrable (giving:
# ln |w+1| - ln|w-1| - 1/2ln|w^2-1| = ln| u| + lnC #
This is now an algebraic problem, and we get:
# ln |w+1|/|w-1| - ln sqrt(w^2-1) = ln |Cu| #
# :. ln( |w+1|/( |w-1|sqrt(w^2-1)) ) = ln |Cu| #
# :. |w+1|/( |w-1|sqrt(w^2-1)) = |Cu| #
And squaring we get:
# (w+1)^2/( (w-1)^2(w^2-1)) = C^2u^2 #
# :. (w+1)^2/( (w-1)^2(w+1)(w-1)) = Au^2 #
# :. (w+1)/( (w-1)^3) = Au^2 #
# :. (w+1) = Au^2 (w-1)^3#
Then restoring the earlier
# v/u+1 = Au^2 (v/u-1)^3#
# :. (v+u)/u = Au^2 ((v-u)/u)^3#
# :. (v+u)/u = Au^2 (v-u)^3/u^3#
# :. v+u = A(v-u)^3#
Finally, we restore the earlier substitutions for
# { (u=x+5/3 ), (v=y-1/3) :} #
Giving us:
# (y-1/3)+(x+5/3) = A((y-1/3)-(x+5/3))^3#
# :. y+x+4/3 = A(y-x-2)^3 #
This is the General Solution, in implicit form.
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Validation of Solutions:
Taking the solution:
# y+x+4/3 = A(y-x-2)^3 #
We have via Implicit Differentiation:
# dy/dx+1 = 3A(y-x-2)^2(dy/dx-1) #
# :. dy/dx+1 = 3A(y-x-2)^2dy/dx-3A(y-x-2)^2 #
# :. (3A(y-x-2)^2-1)dy/dx=3A(y-x-2)^2 +1#
# :. dy/dx = (3A(y-x-2)^2 +1)/(3A(y-x-2)^2-1) #
# \ \ \ \ \ \ \ \ \ \ = (3A(y-x-2)^2 +1)/(3A(y-x-2)^2-1) * (y-x-2)/(y-x-2) #
# \ \ \ \ \ \ \ \ \ \ = (3A(y-x-2)^3 + (y-x-2))/(3A(y-x-2)^3-(y-x-2)) #
# \ \ \ \ \ \ \ \ \ \ = (3(y+x+4/3) + (y-x-2))/(3(y+x+4/3)-(y-x-2)) #
# \ \ \ \ \ \ \ \ \ \ = (3y+3x+4 + y-x-2)/(3y+3x+4-y+x+2) #
# \ \ \ \ \ \ \ \ \ \ = (4y+2x+2)/(2y+4x+6) #
# \ \ \ \ \ \ \ \ \ \ = (2y+x+1)/(y+2x+3) \ \ \ # QED
