Question

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

Answer 1

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)`

Answer 2

(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