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

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

Symbolab says the answer is

##### 2 Answers

# 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

**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

e.g.:

# y = e^(-kt) # ,#y=Acos(omegat )+Bsin(omegat) #

**In the above solution #lnA# was the constant, why?**

We could have used

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

Then we could write

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

There is nothing wrong with this, the solution is still valid,

E.g. If my original question involved constant

# 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

#### Explanation:

Now, when dealing with separable DEs, the objective is to have one side in terms of

So,

We can move

Factor out

We can move

We then integrate both sides.

The integral on the right side looks a bit more complex. It will require partial fraction decomposition of the integrand,

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

Add up the right side.

Equate numerators to find

Let

Let

Thus,

These are simple integrals.

Combine the logarithms to get

The right side is much nicer.

Then, we have

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.

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

Recalling that

Omit the absolute values.

Let

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