Question

Solve the differential equation `(2y-x)dy/dx=2x+y` where `y=3` when `x=2`?

Answer 1

See below.

Explanation:

Making `y = lambda x` we have

`dy = x dlambda + lambda dx` and substituting

`x dlambda + lambda dx = ((2+lambda)/(2lambda-1))dx` or

`x dlambda = (((2+lambda)/(2lambda-1))-lambda)dx` or

`dx/x = -1/2(2lambda-1)/(lambda^2-lambda-1)dy` and after integrating both sides

`log x = -1/2log(2(1+lambda-lambda^2))+C` or

`x = C_0/sqrt(2(1+lambda-lambda^2))` or

`1/2(C_0/x)^2=1+lambda-lambda^2` and solving for `lambda`

`lambda = (x^2 pm sqrt[ 5 x^4-2 C_0^2 x^2])/(2 x^2) = y/x` and finally

`y = x( (x^2 pm sqrt[ 5 x^4-2 C_0^2 x^2])/(2 x^2) ) =`

and simplifying

`y = 1/2(xpmsqrt[ 5 x^2-2 C_0^2])`

For initial conditions we have

`3=1/2(2pm sqrt(20-2C_0^2))` we have

`C_0 = pm sqrt(2)` and finally

`y = 1/2(x pm sqrt(5x^2-4))`

Answer 2

`y^2-xy-x^2+1=0`

Explanation:

We have:

`(2y-x)dy/dx=2x+y`

We can rearrange this Differential Equation as follows:

`dy/dx = (2x+y)/(2y-x)`
`" " = (2x+y)/(2y-x) * (1/x)/(1/x)`
`" " = (2+(y/x))/(2(y/x)-1)`

So Let us try a substitution, Let:

`v = y/x => y=vx`

Then:

`dy/dx = v + x(dv)/dx`

And substituting into the above DE, to eliminate `y`:

`v + x(dv)/dx = (2+v)/(2v-1)`

`:. x(dv)/dx = (2+v)/(2v-1) - v`
`:. " " = {(2+v) - v(2v-1)}/(2v-1)`
`:. " " = {2+v - 2v^2+v}/(2v-1)`
`:. " " = {2+2v - 2v^2}/(2v-1)`
`:. " " = -(2(v^2-v-1))/(2v-1)`

`:. (2v-1)/(v^2-v-1) \ (dv)/dx = -2/x`

This is now a separable DIfferential Equation, and so "separating the variables" gives us:

`int \ (2v-1)/(v^2-v-1) \ dv = - int \ 2/x \ dx`

This is now a trivial integration problem, thus:

`ln(v^2-v-1) = -2lnx + lnA`
`" " = lnA/x^2`

`:. v^2-v-1 = A/x^2`

And restoring the substitution we get:

`(y/x)^2-(y/x)-1 = A/x^2`

Using `y=3` when `x=2`:

`9/4-3/2-1 = A/4 => A = -1`

Thus:

`(y/x)^2-(y/x)-1 = -1/x^2`
`:. y^2-xy-x^2=-1`
`:. y^2-xy-x^2+1=0`