Solving A Nice Differential Equation

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#CalculusProblems #DifferentialEquations

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[randomjin]

To "cheese" through this without any substitution: after multiplying by x you get xy'=y+x², now *differentiate* this to get (xy')'=y'+xy''=y'+2x, so xy''=2x and since x≠0 you get y''=2. Integrate twice, get y=x²+ax+b, substitute back to see that b=0, so finally y=x²+ax

[shmuelzehavi]

Another method:
The given equation is:

f'(x) = f(x)/x + x (1)
Multiply by x:

xf'(x) - f(x) = x^2 (2)
A well known differentiation formula is:

(f(x)/x)' = (xf'(x) - f(x)) / x^2
Therefore from eq. (2) we get:

(f(x)/x)' = 1
Therefore,

f(x)/x = x + C
Or:

f(x) = x^2 + Cx

[sergeykhramov]

I think it would be easier to solve (xy'-y)/x^2 = 1, x is not equal to 0
(y/x)' = 1
y/x = x + C
y = x^2 + Cx

[gniedu]

You could also use the integrating factor, which is perfect for the case y' + P(x)y = Q(x), which is what you have. This would also give that the constant term is 0 right away

[kaveemmeer]

We can try this simple :( xy'-y)/x^2=1
So
Then d(y/x)=dx
Then int{d(y/x)}=int(dx)
Then y/x=x+c
Then y=x^2+cx

[antonyqueen]

It’s easier to differentiate on both sides:
=> f”(x)= [f’(x)/x - f(x)/x^2] + 1
=>f”(x) = 1/x * [f’(x) - f(x)/x] + 1
From the original equation [f’(x) - f(x)/x]=x
Thus we get
f”(x)=2
=> f’(x)=2x + b
=> f(x)=x^2 + bx + c
Using this result in the original equation we get c=0, b any number
=> f(x)= x^2 + bx

[moeberry]

Observe that the differential equation can be rewritten in the form of a first order Linear Differential equation and from there we can find the integrating factor and solve for y at the end. By the way I have used the method of Undetermined coefficients however it was used usually when we have a differential equation with Constant coefficients and not with variable coefficients like in this case. But good to know it can be used in both cases.

[tmsniper]

another method: you can simply divide both sides by x so it becomes
y'/x=y/x²+1
=> y'/x-y/x²=1
notice that we can re-write it as
(y/x)'=1
then it becomes y/x=x+k
multiply both sides by x then you get the result y=x²+kx

[johnchristian]

Nice video, I used the integrating factor method to solve this directly. Good stuff!

[danielbranscombe]

I really like how you show multiple approaches to getting a solution

[amirb]

just divide by x in the beginning y'/x-y/x^2=1 > (y/x)'=1 done!

[holyshit]

First order linear
For particular solution it is better to choose
y = k(x)x because it works for all non-homogeneous first order linear
and also can be generalized do higher orders and also for systems of equations

[jayantsingh]

Ans.
Solution of the given linear differential eqn. is
f(x) = x^2 + c*x
where c is constant.

[satrajitghosh]

After multiply each side by 1/x, a bit of work gives
d/dx ( f(x)/x) = 1
or f(x) = x( x + a)
Here a being a constant of indefinite integration
This represents a parabola

[prexvfn]

I might be wrong or stupid, but I think this equation is easily solved via integrating factor

[cavendishcavendish]

That's great thank you teacher it was my today's maths lesson.

[thierryvieville]

Nice problem again, thanks a lot.

I wonder if we can not avoid the variation of constant method in this case, though this is very powerful, didactic and general method, good to show here.

Very close to the 2nd method it seems we can guess and check ``oh oh y / x here, what about substituting:
z = y / x
here´´, obtaining
z' = y' / x - y / x^2
and since y' = y / x + x
z' = (y / x + x) / x - y / x^2 = 1
so that up to a constant c
z = x + c
yielding
y = x^2 + c x
for any c, easily verifying that it verifies the original equation.

Thank you for all these.

[wristdisabledwriter]

Since I never took differential equations can you show what y h thing looks like

[Youtuber]

It can be solved in one line : (f(x)/x)' =1 !!

[jmart]

I tried this approach
y'=y/x+x
Deriving both sides
y''=(xy'-y)/x^2+1
And substituting y' with the original equation
y''=((y/x+x)x-y)/x^2+1
y''=((y+x^2-y))/x^2+1
y"=2
So
y'=2x+A
y=x^2+Ax+B
And substituting in the original equation you find that B=0
2x+A=2x+A+B/x
So y=x^2+Ax

Thanks again for your nice questions and comments.