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

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

(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