Solve the following differential equations: (i)#(2x+y+3)dy/dx=x+2y+1# and (ii)#[x^(2)+1]dy/dx+2xy=4x^(2)#.?

2 Answers
Apr 14, 2018

Answer:

See below.

Explanation:

Solve the following differential equations:
(i)#(2x+y+3)dy/dx=x+2y+1#.
(ii)#(x^(2)+1)dy/dx+2xy=4x^(2)#.?

i)

Making

#{(u = x+2y+1),(v = 2x+y+3):}#

we have

#{(y=1/3(2u-v+1)),(x = 1/3(2v-u-5)):}#

and also

#{(dy=1/3(2du-dv)),(dx = 1/3(2dv-du)):}#

and also

#dy/dx = (2du-dv)/(2dv-du)# or

#dy/dx = (2-(dv)/(du))/(2 (dv)/(du) - 1) = u/v# or

#(dv)/(du) = (u+2v)/(2u+v)#

now making #v = lambda u rArr (dv)/(du) = (dlambda)/(du) u + lambda #

we have after that new transformation

#(du)/u = (lambda+2)/(1-lambda^2) dlambda# giving

#lnu = 1/2(ln(1+lambda)-3ln(1-lambda))+C_0# then

#u^2 = C_1^2(1+lambda)/(1-lambda)^3# or

#u^2(1-v/u)^3=C_1^2(1+v/u)# or

#(u-v)^3/u=C_1^2(v+u)/u# or

#(u-v)^3=C_1^2(u+v)#

and finally

#(x+y+2)^3+C_1^2(3x+3y+4) = 0# as the general solution for i)

ii)

#(x^2 + 1) y' + 2 x y = 4 x^2#

This is a common linear differential equations with solutions easily obtainable as

#y = 1/(x^2+1)(4/3 x^3+C_0)#

Apr 14, 2018

Answer:

(i) # \ 3y+3x+4 = C(y-x-2)^3#

(ii) # y = (4/3x^3 + C)/(1+x^2) #

Explanation:

ODE (i)

We have:

# (2x+y+3)dy/dx = x+2y+1 #

Which we can write as:

# 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-5/3+2v+2/3+1)/(2u-10/3+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 #v=wu# which if we differentiate wrt #u# using the product gives us:

# (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 #

Then restoring the earlier #w# substitution, using #w=v/u# we get:

# 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#

# :. v+u = A(v-u)^3#

Finally, we restore the earlier substitutions for #u# and #v#, using:

# { (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-1/3-x-5/3)^3#

# :. 3y+3x+4 = C(y-x-2)^3#

Which is the General Solution, in implicit form.

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ODE (ii)

# (x^2+1)dy/dx + 2xy = 4x^2 #

We can write this as:

# dy/dx + (2x)/(1+x^2)y = (4x^2)/(1+x^2) #

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

As the equation is already in this form, then the integrating factor is given by;

# I = exp(int \ P(x) \ dx) #
# \ \ = exp(int (2x)/(1+x^2) \ dx ) #
# \ \ = exp(ln |1+x^2|) #
# \ \ = 1+x^2 #

And if we multiply the DE by this Integrating Factor, #I#, we will end up with the original equation, which we should note was in standard form:

Thus, we can write the equation as a perfect product differential;

# d/dx ((x^2+1)y) = 4x^2 #

This is now separable, so by "separating the variables" we get:

# (x^2+1)y = int \ 4x^2 \ dx #

This is directly integrable, so doing so gives us:

# (x^2+1)y = 4/3x^3 + C #

Leading to the GS:

# y = (4/3x^3 + C)/(1+x^2) #

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Validation of Solutions: ODE (i)

Taking the solution:

# 3y+3x+4 = C(y-x-2)^3#

We have via Implicit Differentiation:

# 3dy/dx+3 = 3C(y-x-2)^2(dy/dx-1)#

# :. dy/dx+1 = C(y-x-2)^2dy/dx-C(y-x-2)^2#

# :. (C(y-x-2)^2-1)dy/dx = C(y-x-2)^2 + 1#

# :. dy/dx = (C(y-x-2)^2 + 1)/(C(y-x-2)^2-1)#

# \ \ \ \ \ \ \ \ \ \ \ = (C(y-x-2)^2 + 1)/(C(y-x-2)^2-1) * (y-x-2)/(y-x-2)#

# \ \ \ \ \ \ \ \ \ \ \ = ((3y+3x+4) + (y-x-2))/((3y+3x+4)-(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

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Validation of Solutions: ODE (ii)

Taking the solution:

# y = (4/3x^3 + C)/(1+x^2) => (1+x^2)y = 4/3x^3 + C #

We have via Implicit Differentiation, and the product rule:

# (1+x^2)dy/dx + (2x)y = 4x^2 \ \ \ \ # QED