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?

2 Answers
Apr 6, 2018

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

Apr 6, 2018

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.

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