Solving the differential equation xy'+2y=x^3

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[kingbeauregard]

The intuition of multiplying by "x" was nifty, but for those who don't just intuit those things, this is a case where you go through the well-known process to create an integrating factor (in this case, "x"). The process is as follows:

1) Divide through by x so that the y' stands alone. So it's y' + (2/x)y = x^2.

2) Integrate the 2/x, which becomes 2ln(x).

3) Your integrating factor will be e^(2ln(x)), or x^2. So take the equation from step 1 and multiply everything by x^2. You get x^2y' + 2xy = x^4.

4) The left side is now "exact" -- the first part is df(x, y)/dy and the second part is df(x, y)/dx -- so the "f(x, y)" that they integrate to is x^2y + C. And the righthand side is x^5/5 (plus its own constant I guess).

[franciscojavierorozcobeiza]

An initial condition should be great, tell me about you please, country ? Bachelor? Occupation?

[holyshit]

First order nonhomogeneous linear 1st method variation of parameter 2nd method integrating factor
Original method used by Leibniz and Bernoulli
Assume that y = uv and plug into equation
x(u'v+uv')+2uv = x^3
xu'v+xuv'+2uv = x^3
xu'v+u(xv'+2v) = x^3
Now we can choose v such that term with u will vanish so
xv'+2v = 0
xv'=-2v
v'/v = -2/x
ln|v|=-2ln|x|
v=1/x^2
u'/x = x^3
u'= x^4
u = x^5/5 + C
y=uv
y = (x^5/5 + C)1/x^2
y=x^3/5+C/x^2
and in my opinion it is old version of variation of parameter method

[angelmendez-rivera]

The domain of y is not specified, so the problem is technically ill-formed, but by convention, one would try to solve an equation such as this on R, and if 0 is in the domain of y and y', then the equation implies y(0) = 0. So with this accounted for, we can safely try to solve for y on R/{0}, and work from there. So we have x·y'(x) + 2·y(x) == x^3 ==> x^2·y'(x) + 2·x·y(x) == x^4 ==> [x^2·y(x)]' = x^4 ==> x^2·y(x) = 1/5·x^5 + A iff x < 0 & x^2·y(x) = 1/5·x^5 + B iff x > 0 ==> y(x) = 1/5·x^3 + A/x^2 iff x < 0 & y(x) = 1/5·x^3 + B/x^2 iff x > 0. Together with the restriction that y(0) = 0, built into the equation itself, we get that y(x) = 1/5·x^3 is the only solution in R, meaning A = B = 0. Normally, this would not happen, since with a specified domain, as a well-formed problem would have, you would not have y(0) = 0 as an inherent restriction by the equation itself. y(x) = 1/5·x^3 with y : R —> R is the only function that is differentiable everywhere and satisfies x·y'(x) + 2·y(x) = x^2 everywhere.

[sngmn]

First of all we have to solve the homogeneous equation in order to find the general solution of given equation:
xy' + 2y = 0
y₀ = Cx⁻²
Then, using the method of variation of an arbitrary constant we find a particular solution:
C'*x⁻¹ = x³ => C' = x⁴ => C = x⁵/5 => y₁ = x³/5
It remains only to add up the particular and general solutions:
y = y₀ + y₁ = Cx⁻² + x³/5. Done.

[dianeweiss]

Cute problem. Thank you for demonstrating that the proposed solution works.

[hans-juergenbrasch]

In the initial diff eq the input of x=0 requires y=0, ie. if 0 is in the domain, then the only solution is x^3/5 and if 0 is not in the domain the general solution is x^3/5+c/x^2.

[daweili]

This is a linear differential equation, which has a standard way to solve.

[niklasmkw]

First solve for y:

xy' +2y = x^3

2y = x^3 - xy'

y = (x^3 - xy')/2

Now notice the highest power we are able to get is x^3, if y is a polynomial equation.
So we get:

y = ax^3 + bx^2 + cx + d
y' = 3ax^2 + 2bx + c

Now plugging those equations into our original equation gives us the following equation:

ax^3 + bx^2 + cx + d = [(1-3a)/2]x^3 - bx^2 - (c/2)x

Setting the coefficients equal to each other gives us a system of equations which is easy to solve:

(1-3a)/2 = a
1-3a = 2a
1 = 5a
a = 1/5

b = -b
2b = 0
b = 0

c = -c/2
2c = -c
3c = 0
c = 0

d = 0 since we dont have a x^0 coefficient on the RHS.

Putting the coefficients a, b, c and d into our polynomial eequation gives us:

y = x^3/5
y' = 3x^2/5

Now we plug those equations into our original equation to check the solution.

x[(3x^2)/5] + 2[(x^3)/5] = x^3
(3x^3)/5 + (2x^3)/5 = x^3
(5x^3)/5 = x^3
x^3 = x^3.

works :)

[InnocentNeuron]

On a preliminary look, I think there may be another shorter way to solve this equation. Dividing by x, we will get a linear equation of the form y' + Py = Q, where P = 2/x and Q = x^2. The integrating function of this equation will be exp(int_(2/x)dx), and the solution will be y = 1/F * int_(Q.F)dx + C, where F is the integrating function.

[ahmadmazbouh]

That was challenging and cool
Keep it up bro <3

[Vladimir_Pavlov]

The "trick" is great. It is based on a specific kind of equation.
But not always the form of the equation allows you to do a "trick". Perhaps it should be pointed out that this equation qualifies as " a linear inhomogeneous differential equation of the first order ." And solve it, in addition to the trick, by some canonical method.
There are two standard (related) algorithms for solving equations of this type.
For me, the Bernoulli method is the most rational.

[dimitris]

you can directly apply the formula for a first order linear differential equation and you can get the same result really easily . Great way of solving it though, reall good explanation

[gelbkehlchen]

Solution:
xy’+2y = x³ ⟹
x*dy/dx+2y = x³ |This is an inhomogeneous, linear differential equation of the first degree, initially the solution of the corresponding homogeneous differential equation ⟹
x*dy/dx+2y = 0 |-2y ⟹
x*dy/dx = -2y |*dx/(x*y) ⟹
dy/y = -2*dx/x |∫() ⟹
∫dy/y = -2*∫dx/x ⟹
ln|y| = -2*ln|x|+C |e^() ⟹
y = e^(-2*ln|x|+C) = e^C*x^(-2) = K*x^(-2) = K/x² |The solution of the original, inhomogeneous differential equation with variation of the constants. ⟹
y = K(x)*x^(-2) ⟹ y’ = K’(x)*x^(-2)-2*K(x)*x^(-3) | inserted into the original, inhomogeneous differential equation ⟹
= x³ ⟹
= x³ ⟹
K’(x)*x^(-1) = x³ |*x ⟹
K’(x) = x^4 |∫()*dx ⟹
K(x) = x^5/5+C1 ⟹
The whole solution is then:
y = K(x)*x^(-2) = (x^5/5+C1)*x^(-2) = x³/5+C1/x²

Sample by substituting it into the differential equation:
y’ = 3/5*x²-2*C1/x³ ⟹
Left side: xy’+2y =
= 3/5*x³-2*C1/x²+2/5*x³+2*C1/x² = x³
Right side: x³ everything ok.

[math]

very well explained, thanks for sharing this solution

[majidmelek]

What About or using the Laplace transform?

[wphhvjb]

I think that try to multiply by x is difficurt.
It would have been easy if We had thought of it !!
And it's just a simple integral.

[armacham]

I'll share what I did. I can't say it's a better method. I just took the derivative over and over again until RHS was zero.

xy'' + 3y' = 3xx

xy''' + 4y'' = 6x

xy'''' + 5y''' = 6

+ 6y'''' = 0


re-arrange it to get: = -6/x


​integrate both sides to get ln(abs(y'''')) = -6ln(abs(x)) + c


replace c with ln(c2)

ln(abs(y'''')) = -ln(x^6) + ln(c2) = ln(c2/x^6)



​e both sides to get abs(y'''') = c2/x^6


y'''' = +-c2/x^6

c3 = +-c2

y'''' = c3/x^6

now integrate over and over again and keep replacing the "c"s to get rid of coefficients

y''' = -5c3/x^5 + c4


y''' = c5/x^5 + c6

y'' = -4c5/x^4 + x*c6 + c7

y'' = c8/x^4 + x*c6 + c7

y' = -3c8/x^3 + x^2*c6/2 + x*c7 + c9

y' = c10/x^3 + c11*x^2 + c7*x + c9

y = -2c10/x^2 + c11/3*x^3 + c7*x^2/2 + c9*x + c12

y = c13/x^2 + c14*x^3 + c15*x^2 + c16*x + c17

replace the c terms with letter variables for simplicity

y = ax^3 + bx^2 + cx + d + e/x^2

now go back to the original equation in order to solve for the values for those variables a, b, c, d, e

take the derivative of this new equation

y' = 3ax^2 + 2bx + c - 2e/x^3


xy'+2y=x^3

x*(3ax^2 + 2bx + c - 2e/x^3) + 2*( ax^3 + bx^2 + cx + d + e/x^2) = x^3

3ax^3 + 2bx^2 + cx - 2e/x^2 + 2ax^3 + 2bx^2 + 2cx + 2d + 2e/x^2 = x^3

combine like terms. notice -2e and e2, those cancel out.

5ax^3 + 4bx^2 + 3cx + 2d = x^3

from here it should be obvious that b, c, and d must be equal to zero

5ax^3 = x^3

And that a = 1/5

go back and plug those values in, you get:

y = x^3/5 + e/x^2

which is the answer, where "e" is a variable representing any real number. for the sake of simplicity let's replace "e" with "c"

y = x^3/5 + c/x^2
| c is a member of real numbers

[jimschneider]

I did it the slow way (spoiler alert - hit "read more" to see it).




First, I noticed that if y = a x^n, then x y' = n y, so x y' + 2 y = (n+2) y = (n+2) a x^n. Plugging this into the equation gave (n+2) a x^n = x^3, so n = 3, and a = 1/(3+2) = 1/5, making y = x^3/5.

Next, if z is a solution to x z' + 2 z = 0, then any multiple of z can be added to y above, and the result would still be a solution to the original equation. As it happens, x z' + 2 z = 0 is separable, and gives the equation ln(z) = - 2 ln(z) + k, or z = C/x^2. This means the complete, general solution is y = x^3/5 + C/x^2 (exactly what SyberMath got in a more clever way).

[anshumanagrawal]

Kudos to you for not using any "well known formulas" and actually explaining your solution, not assuming that your viewers are just already familiar with differential equations