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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