Question

Solve `(x - 2y + 1)dx + (4x - 3y - 6)dy = 0` and give an explicit solution?

Answer 1

`:. (3y+x-9)^5=C(y-x+1),` is the desired GS.

Explanation:

We have, `dy/dx+(x-2y+1)/(4x-3y-6)=0....(ast).`

Let, `X=x-h, Y=y-k, h,k" constants."`

`:. dx=dX, and, dy=dY, or, dy/dx=(dY)/(dX).`

`:. (ast)" becomes, "(dY)/(dX)+(X-2Y+h-2k+1)/(4X-3Y+4h-3k-6)=0.`

Choose `h,k` such that,

`h-2k+1=0=4h-3k-6, i.e., h=3, k=2.`

`:.X=x-3, &, Y=y-2rArr(dY)/(dX)+(X-2Y)/(4X-3Y)=0....(ast').`

This is a Homogeneous Diff. Eqn., &, to find its General Solution (GS),

we substitute `Y=VXrArr (dY)/(dX)=V+X(dV)/(dX).`

`:.(ast')rArr V+X(dV)/(dX)+(X-2VX)/(4X-3VX)=0, or,`

`X(dV)/(dX)+(1-2V)/(4-3V)+V=0, i.e., X(dV)/(dX)+(1+2V-3V^2)/(4-3V)=0`.

`:.(3V-4)/(3V^2-2V-1)dV+1/XdX=0...."[Separable Variable]."`

`:.int(3V-4)/(3V^2-2V-1)dV+int1/XdX=lnc.`

`:.int[1/4{15/(3V+1)-1/(V-1)}]dV+lnX=lnc.`

`:.1/4{15*1/3ln(3V+1)-ln(V-1)}+lnX=lnc.`

`:.5/4ln(3V+1)-1/4ln(V-1)+lnX=lnc.`

`:.5ln(3V+1)-ln(V-1)+4lnX=4lnc=lnC.....[C=c^4].`

Since, `V=Y/X=(y-2)/(x-3),` we have,

`5ln{(3(y-2))/(x-3)+1}-ln{(y-2)/(x-3)-1}+4ln(x-3)=lnC.`

`:.5ln{(3y+x-9)/(x-3)}-ln{(y-x+1)/(x-3)}+4ln(x-3)=lnC.`

`:.ln{(3y+x-9)/(x-3)}^5-ln{(y-x+1)/(x-3)}+ln(x-3)^4=lnC.`

`:.ln[{(3y+x-9)/(x-3)}^5xx{(x-3)/(y-x+1)}(x-3)^4]=lnC.`

`:.ln[(3y+x-9)^5/{(y-x+1)}]=lnC.`

`:. (3y+x-9)^5=C(y-x+1),` is the desired GS.

Enjoy Maths.!

Answer 2

See below.

Explanation:

Making the change of variables

`((1,-2),(4,-3))((x),(y))=((u),(v))`

we have

`(u+1)(-3du+2dv)+(v-6)(-4du+dv) = 0`

Making now

`{(U=u+1),(V = v-6):}` we have

`U(-3dU+2dV)+V(-4dU+dV)= 0` or

`-3UdU+2U dV-4V dU+V dV = 0`

introducing now the change of variables

`{(eta = U^2),(xi = V^2):}` we have

`-3/2d eta+1/2 d xi +sqrt(eta/xi) d xi-2 sqrt(xi/eta)d eta = 0` or

`(1/2+sqrt(eta/xi))d xi = (3/2+2sqrt(xi/eta)) d eta`

Making now the change of variables

`eta = lambda xi rArr d eta = lambda d xi + xi d lambda`

`(1/2+sqrt lambda)d xi= (3/2+2/sqrt(lambda))(lambda d xi + xi d lambda)`

and now calling `f(lambda) = ((1/2+sqrt lambda)/ (3/2+2/sqrt(lambda)))` we have now

`(f(lambda)-lambda)d xi = xi d lambda` which is separable

`(d lambda)/(f(lambda)-lambda)=(d xi)/xi` giving

`-5/2 log(1 - 3 sqrt lambda) + 1/2 log(1 + sqrt lambda)= log xi + C_0` or

`(1+sqrtlambda)/(1-3sqrtlambda)^5 = C_0^2 xi^2`

The recovery to the `x, y` variables is left as an exercise.

Answer 3

`y-x+1 = (3y+x-9)^5`

Explanation:

We have:

`(x - 2y + 1)dx + (4x - 3y - 6)dy = 0`

Which we can write as:

`dy/dx = - (x - 2y + 1)/(4x - 3y - 6)` ..... [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 ), (4x - 3y - 6=0) :} => { ( x=3 ), (y=2) :}`

As a result we perform two linear transformations:

Let `{ (u=x-3 ), (v=y-2) :} => { ( x=u+3 ), (y=v+2) :}`

And if we substitute into the DE [A] we get

`(dv)/(du) = - ((u+3) - 2(v+2) + 1)/(4(u+3) - 3(v+2) - 6)`
`\ \ \ \ \ = - (u+3 - 2v-4 + 1)/(4u+12 - 3v-6 - 6)`
`\ \ \ \ \ = - (u- 2v)/(4u- 3v)` .... [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)/(4u- 3wu)`

`:. w + u(dw)/(du) = - (1- 2w)/(4- 3w)`

`:. u(dw)/(du) = - (1- 2w)/(4- 3w) -w`

`:. u(dw)/(du) = ( (2w-1) -w(4- 3w) )/(4- 3w)`

`:. u(dw)/(du) = ( 3w^2-2w-1 )/(4- 3w)`

This is now a separable DE, so we can rearrange and separate the variables to get:

`int \ (4- 3w)/( 3w^2-2w-1 ) \ dw= int \ 1/u \ du`

The denominator in the LHS factorises and we can decompose into partial fractions (omitted) giving

`int \ 1/(4(w-1)) -15/(4(3w+1)) \ dw= int \ 1/u \ du`

Which is now readily integrable giving:

`1/4 ln (w-1) -5/4ln(3w+1) = ln u + lnA`

`:. ln (w-1) -ln(3w+1)^5 = 4ln Au`

`:. ln (w-1)/(3w+1)^5 = ln Cu^4`

`:. (w-1)/(3w+1)^5 = Cu^4`

Then restoring the earlier substitution we have:

`w=v/u = (y-2)/(x-3)`

Thus:

`((y-2)/(x-3)-1)/(3(y-2)/(x-3)+1)^5 = C(x-3)^4`

We can simplify by placing each fraction over a common denominator:

`( ( (y-2)-(x-3))/(x-3) )/( (3(y-2)+x-3)/(x-3))^5 = C(x-3)^4`

`:. ( ( y-2-x+3)/(x-3) ) * ( (x-3) / (3y-6+x-3))^5 = C(x-3)^4`

`:. ( (y-x+1) * (x-3)^4 ) / (3y+x-9)^5 = C(x-3)^4`

`:. y-x+1 = (3y+x-9)^5`

This is the General Solution, and we would struggle to get an explicit solution