I try to solve this by differentiating. To shorten the notation, I call y0(x), y1(x), y2(x), y3(x) the function itself, its first, second, third derivative and so on. 2nd derivative y2: y2(x) = y0(x) + x^3 Take the next derivatives on both sides: 3rd derivative y3: y3(x) = y1(x) + 3x^2 4th derivative y4: y4(x) = y2(x) + 6x 5th derivative y5: y5(x) = y3(x) + 6 6th derivative y6: y6(x) = y4(x) Set z0(x) = y1(x) z1(x) = y2(x) z2(x) = y3(x) z3(x) = y4(x) z4(x) = y5(x) z5(x) = y6(x) Then we get the ODE z5(x) = z3(x) It is known from physics (harmonic oscillations) that if a function shall be the same as its second derivative, this ODE can be solved by a sine or cosine function, e.g. s(t) = s_max_ * sin(omega * t + phi) with t = time, s_max_ = amplitude and phi = phase shift of the harmonic oscillation. So we could use the ansatz z3(x) = a * sin(b*x + c) + d And integrate this function: y3(x) = z2(x) = (a /b) * (- cos(b*x + c)) + d*x + e y2(x) = z1(x) = (a /b^2) * (- sin(b*x + c)) + d/2*x^2 + e*x + f y1(x) = z0(x) = (a /b^3) * ( cos(b*x + c)) + d/6*x^3 + e/2*x^2 + f*x + g y0(x) = (a /b^4) * ( sin(b*x + c)) + d/24*x^4 + e/6*x^3 + f/2*x^2 + g*x + h So our searched function class is y(x) = (a/b^4) * sin(b*x + c) + d/24*x^4 + e/6*x^3 + f/2*x^2 + g*x + h What we now need is: y'(x) = (a/b^3) * cos(b*x + c) + d/6*x^3 + e/2*x^2 + f*x + g y''(x) = (-a/b^2) * sin(b*x + c) + d/2*x^2 + e*x + f We want y''(x) = y(x) + x^3. To achieve this, we must assure a/b^4 = -a/b^2 That means a = 0 or b^4 = -b^2 Since b = 0 is forbidden because b is in the denominator, we can divide by b: b^2 = -1 and must allow b = +- i (complex values for b). And we must assure d/24*x^4 = 0*x^4 d/24 = 0 d = 0 e/6*x^3 + x^3 = 0*x^3 e/6 + 1 = 0 e/6 = -1 e = -6 f/2*x^2 = d/2*x^2 f/2 = d/2 f = d = 0 g*x = e*x g = e = -6 h = f = 0 Altogether, I get the following solution functions: for a = 0, a cubic function: y(x) = - x^3 - 6x for b^2 = -1, a complex sine function together with a cubic function: y(x) = (-a) * sin((+-i)*x + c) - x^3 - 6x (The latter case also comprises the first case).
Question on the homog vs particular solution. What you provided in the 1st solution is the standard method that's taught in diff eq. However, if I hadn't studied that, and if I was just trying to solve this problem not knowing better, I would simply use a polynomial as you were doing in the Particular solution. Not to say that would be correct. This is probably a dumb question, but why do we always include the homogeneous solution?
Remember, we're aiming to find ALL solutions that make the given differential equation true. Yes, if we just took the yp, found the coefficients (a,b,c,d) = (-1,0,-6,0), so that yp = - x^3 - 6x, which is of course a solution. But, we see that this solution yp is one that's included in SyberMath's general solution of y = c1 e^x + c2 e^(-x) - x^3 - 6x , provided we let c1 = c2 = 0. But are there other function that are also solutions to the given diff. eqn? Yes, an infinite number of them! There's one if you let c1 = 1 and c2 = 6. Another, if you let c1 = 0 and c2 = pi/12, etc... If fact, you get another distinct solution just by picking c1 & c2 as any pair of real numbers! That's why c1 & c2 are "arbitrary constants"; IT DOESN'T MATTER what c1 & c2 are, when plugged into the general solution y, it will give a true statement when plugged into the differential equation. That's why the y obtained is called a "general solution". ALL solutions to the diff eqn are one of the ones given by that solution. Hope this helps.
it's enough to plug the ansatz for the particular solution into the equation to get the coefficients. the homogeneous solution cancels anyway which is the point of solving the homogeneous problem first
We can also do yp = x^3 / (D^2 -D)? Then yp = x^3 / D(D-1) which is 1/D ( D-1)^-1 x^3 Which ultimately becomes -1/D (x^3 + 6x + 6) which integrates to -x^4 / 4 + 3x^2 +6x?
The particular solution itself without homogenous solution satisfy the differential equation so it is enough to plug the yp and we get the value of a,b,c and d add yh
That's what I wondered. The method he used for y looked like Laplace. You would get s^2 Y - sy(0) - y(0) = Y + 6/s^4 Y = [ 6/s^4 + sy(0) + y(0) ] / (s^2 - 1) Does the inverse Laplace give the answer? Not sure I have the motivation to solve it!
40 years after coming across these things I think I finally understand them!
I try to solve this by differentiating. To shorten the notation,
I call y0(x), y1(x), y2(x), y3(x) the function itself, its first, second, third derivative and so on.
2nd derivative y2: y2(x) = y0(x) + x^3
Take the next derivatives on both sides:
3rd derivative y3: y3(x) = y1(x) + 3x^2
4th derivative y4: y4(x) = y2(x) + 6x
5th derivative y5: y5(x) = y3(x) + 6
6th derivative y6: y6(x) = y4(x)
Set
z0(x) = y1(x)
z1(x) = y2(x)
z2(x) = y3(x)
z3(x) = y4(x)
z4(x) = y5(x)
z5(x) = y6(x)
Then we get the ODE
z5(x) = z3(x)
It is known from physics (harmonic oscillations)
that if a function shall be the same as its second derivative,
this ODE can be solved by a sine or cosine function, e.g.
s(t) = s_max_ * sin(omega * t + phi)
with t = time, s_max_ = amplitude and phi = phase shift of the harmonic oscillation.
So we could use the ansatz
z3(x) = a * sin(b*x + c) + d
And integrate this function:
y3(x) = z2(x) = (a /b) * (- cos(b*x + c)) + d*x + e
y2(x) = z1(x) = (a /b^2) * (- sin(b*x + c)) + d/2*x^2 + e*x + f
y1(x) = z0(x) = (a /b^3) * ( cos(b*x + c)) + d/6*x^3 + e/2*x^2 + f*x + g
y0(x) = (a /b^4) * ( sin(b*x + c)) + d/24*x^4 + e/6*x^3 + f/2*x^2 + g*x + h
So our searched function class is
y(x) = (a/b^4) * sin(b*x + c) + d/24*x^4 + e/6*x^3 + f/2*x^2 + g*x + h
What we now need is:
y'(x) = (a/b^3) * cos(b*x + c) + d/6*x^3 + e/2*x^2 + f*x + g
y''(x) = (-a/b^2) * sin(b*x + c) + d/2*x^2 + e*x + f
We want y''(x) = y(x) + x^3.
To achieve this, we must assure
a/b^4 = -a/b^2
That means a = 0 or
b^4 = -b^2
Since b = 0 is forbidden because b is in the denominator,
we can divide by b:
b^2 = -1
and must allow b = +- i (complex values for b).
And we must assure
d/24*x^4 = 0*x^4
d/24 = 0
d = 0
e/6*x^3 + x^3 = 0*x^3
e/6 + 1 = 0
e/6 = -1
e = -6
f/2*x^2 = d/2*x^2
f/2 = d/2
f = d = 0
g*x = e*x
g = e = -6
h = f = 0
Altogether, I get the following solution functions:
for a = 0, a cubic function: y(x) = - x^3 - 6x
for b^2 = -1, a complex sine function together with a cubic function:
y(x) = (-a) * sin((+-i)*x + c) - x^3 - 6x
(The latter case also comprises the first case).
Question on the homog vs particular solution. What you provided in the 1st solution is the standard method that's taught in diff eq. However, if I hadn't studied that, and if I was just trying to solve this problem not knowing better, I would simply use a polynomial as you were doing in the Particular solution. Not to say that would be correct.
This is probably a dumb question, but why do we always include the homogeneous solution?
Remember, we're aiming to find ALL solutions that make the given differential equation true.
Yes, if we just took the yp, found the coefficients (a,b,c,d) = (-1,0,-6,0), so that yp = - x^3 - 6x, which is of course a solution.
But, we see that this solution yp is one that's included in SyberMath's general solution of
y = c1 e^x + c2 e^(-x) - x^3 - 6x , provided we let c1 = c2 = 0.
But are there other function that are also solutions to the given diff. eqn? Yes, an infinite number of them!
There's one if you let c1 = 1 and c2 = 6. Another, if you let c1 = 0 and c2 = pi/12, etc...
If fact, you get another distinct solution just by picking c1 & c2 as any pair of real numbers!
That's why c1 & c2 are "arbitrary constants"; IT DOESN'T MATTER what c1 & c2 are,
when plugged into the general solution y, it will give a true statement when plugged into the differential equation.
That's why the y obtained is called a "general solution". ALL solutions to the diff eqn are one of the ones given by that solution.
Hope this helps.
@@timeonly1401 it does. Thanks!
it's enough to plug the ansatz for the particular solution into the equation to get the coefficients. the homogeneous solution cancels anyway which is the point of solving the homogeneous problem first
We can also do yp = x^3 / (D^2 -D)?
Then yp = x^3 / D(D-1) which is 1/D ( D-1)^-1 x^3
Which ultimately becomes -1/D (x^3 + 6x + 6) which integrates to -x^4 / 4 + 3x^2 +6x?
Would you (please) do a video to explain and prove where the idea of characteristic equations comes from?
The particular solution itself without homogenous solution satisfy the differential equation so it is enough to plug the yp and we get the value of a,b,c and d add yh
Nice!
Thanks!
Is it possible to solve this equation using Laplace transform?
That's what I wondered. The method he used for y looked like Laplace.
You would get
s^2 Y - sy(0) - y(0) = Y + 6/s^4
Y = [ 6/s^4 + sy(0) + y(0) ] / (s^2 - 1)
Does the inverse Laplace give the answer? Not sure I have the motivation to solve it!
After you solved the first part I solved the second part. I hope It is correct. It took me two anda half house, I found It Very challenging.
@@Guilles-z4d It looked bad! I presume partial fractions.
I don't know which grade math is this but you surely can use Laplace transform for this.
yp=-x^3-6x..yo=c1e^x+c2e^(-x)
Your 2b or not 2b makes my skin burn. Your videos would otherwise be useful.