Question

What is the general solution of the differential equation `y-xy'=3-2x^2y'`?

Symbolab says the answer is `y=(xc_1)/(2x-1)+3`?

Answer 1

`y = 3+(Cx)/(2x-1)`

Explanation:

We have:

`y-xy'=3-2x^2y'`

Which we can rearrange and collect terms to get:

`2x^2y'-xy' + y =3`

`:. (2x^2-x)y' =3-y`

`:. 1/(3-y) dy/dx = 1/(x(2x-1))`

Which is a separable ODE, so if we "separate the variables" then:

`- int \ 1/(y-3) \ dy = int \ 1/(x(2x-1)) \ dx`

We can decompose the RHS integral into partial fractions, and we gain:

`- int \ 1/(y-3) \ dy = int \ 2/(2x-1)-1/x \ dx`

And integrating we get:

`- ln(y-3) = ln(2x-1)-lnx + lnA`

`:. ln(1/(y-3)) = ln ((A(2x-1))/x)`

`:. 1/(y-3) = (A(2x-1))/x`

`:. y-3 = x/(A(2x-1))`

`:. y = 3+(Cx)/(2x-1) \ \ \ \` where `C=1/A \ \ \ \` QED

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Side Note on the Constant of Integration:

In order to answer the follow-up query where to did `ln A` come from, it is is easier to explain in the solution in FAQ form.

Why do we have (or need) a constant in a DE solution?

Quite simply because the derivative of any constant is zero, then when we integrate (which we do solve a DE) there is always the possibility that the original DE solution contained a constant,. eg:

`{ (y=2x+1), (y=2x+2), (y=2x+c) :} => dy/dx=2`

So the "General" solution of the DE:

`dy/dx=2 " is " y=2x + C`

Every order of the DE will introduce a constant, so a Second Order equation will introduce two (independent) constants:

Why is the constant always called `C`?

It isn't - it can be called `C`, `A`, `k` or anything else suitable. It is just a mathematical symbol used to denote a constant. commonly it is called `C` (for constant, as is frequently used when we perform an integration) but equally common alternatives vary according to the solution, eg: `k` for exponential decay `A,B` for sinoisiodal solutions

e.g.: `y = e^(-kt)`, `y=Acos(omegat )+Bsin(omegat)`

In the above solution `lnA` was the constant, why?

We could have used `C` and gained the solution

`- ln(y-3) = ln(2x-1)-lnx + C`

Then we could write `C=ln e^C` (using the rules of logs), so the final solution would be

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

There is nothing wrong with this, the solution is still valid, `1/e^C` is still a constant but by writing the constant in a different form we can gain a simpler solution. In this case seeing that all terms involved a logarithm, then writing the constant as the log of a constant simplified things, this is a common technique.

E.g. If my original question involved constant `alpha` and `beta` and upon solving the DE we used `C` is the constant and gained a solution of the form:

`y = alphacos(omegat )+betasin(omegat) + (tan(alpha+beta))/(cosalpha+sinbeta) + alpha + beta + C`

We can more concisely write this as:

`y = alphacos(omegat )+betasin(omegat) + A`

You will often encounter a solution whose solution itself generates a constant term, and this constant then mysteriously vanishes because it is incorporated into the constant that came form integration

Answer 2

`y=3-(Ax)/(2x-1)`

Explanation:

Now, when dealing with separable DEs, the objective is to have one side in terms of `y, dy` and the other in terms of `x, dx`. This one looks a bit rough to separate, but after a few steps, it's not so bad.

So,

`y-xy'=3-2x^2y'`

We can move `2x^2y'` over to the left.

`2x^2y'-xy'=3-y`

Factor out `y'.`

`y'(2x^2-x)=3-y`

We can move `2x^2-x` to the right, entailing division by it on the right, and `3-y` to the left, entailing division by it on the left. Also, replace `y'` with `dy/dx`.

`dy/dx(1/(3-y))=1/(2x^2-x)`

`dy/(3-y)=dx/(2x^2-x)`

We then integrate both sides.

`intdy/(3-y)=intdx/(2x^2-x)`

The integral on the right side looks a bit more complex. It will require partial fraction decomposition of the integrand, `1/(2x^2-x)`

Factor the denominator and decompose in accordance with the rules for decomposition involving linear factors in the denominator.

`1/(x(2x-1))=A/x+B/(2x-1)`

Add up the right side.

`1/(x(2x-1))=(A(2x-1)+Bx)/((x)(2x-1))`

Equate numerators to find `A,B.`

`1=A(2x-1)+Bx`

Let `x=0.`

`1=-A, A=-1`

Let `x=1/2.`

`1=1/2B, B=2`

Thus, `intdx/(x(2x^2-1))=-intdx/x+2intdx/(2x-1)`

These are simple integrals.

`-intdx/x+1/2intdx/(2x-1)=-ln|x|+ln|2x-1|`

Combine the logarithms to get `ln|(2x-1)/x|`

The right side is much nicer. `intdy/(3-y)=-ln|3-y|`

Then, we have

`-ln|3-y|=ln|(2x-1)/x|+C`

Do not forget that constant, it is important. We would technically have two constants, one from each side, but we absorbed them into the single constant on the right.

`ln|3-y|=-ln|(2x-1)/x|+C`

This implicit solution is not good enough, we need an explicit solution. Exponentiate both sides.

`e^(ln|3-y|)=e^((-ln|(2x-1)/x)|+C)`

Recalling that `e^lna=a, e^(a*b)=e^ae^ b`:

`|3-y|=e^(-ln|(2x-1)/x|)*e^C`

`-ln|(2x-1)/x|=ln|((2x-1)/x)^-1|=ln|x/(2x-1)|`

`|3-y|=e^(ln|x/(2x-1)|)*e^C`

`|3-y|=|x/(2x-1)|e^C`

Omit the absolute values.

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

Let `A=+-e^C`, this is just an arbitrary constant.

`3-y=(Ax)/(2x-1)`

`y=3-(Ax)/(2x-1)`

This does fall in line with the answer Symbolab gave me.