A Differential Equation WA Cannot Solve?
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Inverse of a quadratic is also a possible solution. Something if the sort 1 / (a.x2 + b.x + c). Because in this case y'' becomes something like 1 / (p.x^4 + q.x^3 + r.x^2 + s). Dividing thid by y, you end up with a quadratic which can be equated to 4.x^3 + 2 and we can find values for a, b, c, p, q, r and s.
Why should y'' have 1 in the numerator? Even in that case, if you divide 1 / (p.x^4 + q.x^3 + r.x^2 + s) by 1 / (a.x2 + b.x + c), you obtain an algebraic fraction in which the numerator has degree 2, the denominator has degree 4, which, of course, is not a polynomial.
Alternative method. Multiply the original equation by y. (This introduces the constant function 0 as an extraneous solution.) The result is a homogeneous second order linear ODE with non-constant coefficients. This can be solved completely by series methods.
In order to find the full solution, you seek it in the form y=ve^u, where v and u are independent functions of x. Plugging y into the original equation, we obtain (v’’+2v’u’)/v+(u’’+(u’)^2)=4x^2+2. Using the particular solution found in the video for u=x^2, we obtain that (v’’+2v’u’)/v=0 or v’’+2v’u’=0 and v’’+4xv’=0, the last being an ordinary second order linear differential equation with respect to v with the solution v=A⎰e^(-2x^2)dx + B (A and B - constants) Thus, the full solution is y=(A⎰e^(-2x^2)dx + B)e^(x^2)
Your result checks out!🙂
You're good at this 👍
You forgot the c
It's zero
@shogun6943 c is not 0, it can be any number
Yes mb
It's just a constant
He actually forgot 2 constants
I'll just point out that 4x^2-2 (note minus sign) is the second-order Hermite polynomial, which is the second derivative of a Gaussian divided by the original function. Hermite polynomials arise in the quantum mechanical wavefunctions of a harmonic oscillator.
Cool!
y = Ae^(x^2 + C) is a family of solutions with apparently two parameters generally required when discussing second-order differential equations. Indeed, the constant C is useless, because it can be absorbed into the arbitrary constant A.
A few boring observations of mine :D
So, we have a 2ⁿᵈ order homogeneous liner ordinary differential equation (LODE) with variable coefficients of a type: y’’(x) = a(x)·y’(x) + b(x)·y(x), where a(x)=0 and b(x) = 4x² +2.
Theory says that we MUST! have 2 and only 2 linearly independent particular solutions y₁(x) and y₂(x) of this ODE and the general solution is y(x) = C₁·y₁(x) + C₂·y₂(x), C₁ and C₂ can be found through initial or boundary conditions. The problem here is though, that we don't have an apparatus at hand to derive solutions for such 2ⁿᵈ order ODE with variable coefficients in a systematic way. There's however, a very useful observation related to Wronskian and Liouville formula.
For solutions y₁(x) and y₂(x), we can, by definition, determine Wronskian as W(x) = y₁(x)·y₂'(x) - y₂(x)·y₁'(x). Then, it is pretty easy to find that W'(x) = a(x)·W(x) [if you differentiate, substitute y''(x) = a(x)·y'(x) + b(x)·y(x) and note that some terms will cancel out]. Since a(x) = 0 in our case, W'(x) = 0 and W(x) = const = C₀. Note that C₀≠ 0 for any x, as Wronskian is always non-zero for any linearly independent functions y₁(x) and y₂(x).
But then it means that, W(x) = y₁(x) ·y₂'(x) - y₂(x)·y₁'(x) = C₀.
So, if we are lucky and guess y₁(x), then this formula enables us to find y₂(x) using this 1ˢᵗ order ODE, that we do know how to solve. And this is exactly the case here!
As you showed, we can find by smart guessing that y₁(x) = exp(x²). Then, y₁'(x) = 2x·exp(x²).
Then, W(x) = y₁(x)·y₂'(x) - y₂(x)·y₁'(x) = [ y₂'(x) - 2x·y₂(x) ] · exp(x²) = C₀ or
y₂'(x) - 2x · y₂(x) = C₀ · exp(-x²)
We can now start looking for the solution of this non-homogeneous 1st order ODE with variable coefficients in order to find y₂(x). This is a bit cumbersome, but quite straight-forward. If we temporarily put the RHS equal to 0 to find the solution of related homogeneous ODE, we will find homogeneous solution:
y₂(x)ʰ = A·exp(x²).
We then seek the general solution of this 1st order non-homogeneous ODE (with a proper non- zero RHS [C₀·exp(-x²)] as before) using the method of variation of parameters:
y₂(x) = A(x) · exp(x²) --substitute into the [ y₂'(x) - 2x · y₂(x) = C₀ · exp(-x²) ]
A'(x) · exp(x²) + -A(x)·2x·exp(x²)- - -2x·A(x)·exp(x²)- = C₀·exp(-x²) --divide by exp(x²)
A'(x) = C₀ · exp(-2x²)
A(x) = C₀ · ∫ exp(-2x²) dx + B
we can now put u = x√2, then x = u· (1/√2), dx= du· (1/√2) and
A(x) = C₀ · (1/√2) · ∫ exp(-u²) du + B = C₀ · (1/2) · √(π/2) · (2/√π) · ∫ exp(-u²) du + B
The function [ (2/√π)·∫ exp(-u²) du ] is called "the error function" erf(u). In fact, to be rigorous the integral would need to be definite (not indefinite as now) and go from 0 to some x, but this is known to be just some constant (let's say "D") away from an indefinite integral ∫ ... du. As we already have an unknown constant B, then we can just re-define B as B+D, and we will be fine then. So, we can safely say that:
A(x) = C₀ · (1/2) · √(π/2) · erf(x√2) + B ---remember that u = x√2
y₂(x) = A(x) · exp(x²) = C₀ · (1/2) · √(π/2) · erf(x√2) · exp(x²) + B · exp(x²)
So we just found the y₂(x).
Then, as y(x) = C₁·y₁(x) + C₂·y₂(x):
y(x) = C₁ · exp(x²) + C₂ · [ C₀ · (1/2) · √(π/2) · erf(x√2) · exp(x²) + B · exp(x²) ]
Re-group and obtain:
*y(x) = exp(x²) · [ D₁ + D₂ · erf(x√2) ]*
where we re-defined the constants as:
D₁ = C₁ + C₂ · B
D₂ = C₀ · C₂ · 1/2 · √(π/2)
Note that this approach guarantees that there cannot be any other linearly independent solutions of the initial 2nd order ODE.
Setting t'=u(x) we get u^2+u'=(2x)^2+2 so u^2=(2x)^2 and u'=2 so u= + - 2x AND u'=2 so u=+2x. Also t'=u=2x so t=x^2+c hence y=e^t=e^(x^2+c)=e^(x^2)*e^c. Setting e^c=C we get y=C*e^(x^2)
Good question: Can you integrate y=e^(x^2)? 🤩
From a system of differential equations, you have (dt/dx)²=4x² and d²t/dx²=2, then you can intercept the solutions and see the general form that works 💪
You had (t')^2+t"=4x^2+2. Substitute t'=(2x)^2,then t"=2, then integral of t=x^2. Hence y=e^(x^2)
😊🎉👍👍👍🎉😊
Apparently D_n of z is the Parabolic Cylinder Function, never heard of it
Wow, nice one
y = a*exp(x^2 + b*x + c)
Sorry, b must be 0, so y = a*exp(x^2).
@@stephenshefsky5201 You mean y = e*exp(x^2)
+c@@MaxCubing11
problem
y"/y = 4x² + 2
Note that y cannot be 0.
By guess and check:
if
y = ae^(x²)
, where a is a real constant,
then
y' = 2ax e^(x²)
y" = ae^(x²) (4x²+2)
y"/y = ae^(x²) (4x²+2)/ae^(x²) (4x²+2)
= 4x² +2
Therefore one solution is
y = a e^x²
Alternatively,
Let p= dy/dx
y = ∫ p dx
1/∫ p dx = (4x² +2) dx/dp
∫ p dx = dp/dx (1/(4x² +2))
Take derivative.(this makes it a third order equation)
[(4x² +2) d²p/dx²- 8x dp/dx] / [(4x²+2)²] = p
, where p = dy/dx.
Now the equation has transformed to the third order
d³y/dx³ -(4x/(2x²+1)) d²y/dx²- (4x²+2)dy/dx = 0
with solution
y = C₁ e^(x²+½)+√(π/2) C₂ e^(x²-½) erf(√2 x)
We see that
g=e^(x²-½) erf(√2 x)
is linearly independent of y = e^(x²+½).
WA gives
d(erf(x))/dx = 2 e^(-x²) /√π
, very helpful in showing that
g(x) = √(π/2) e^(x²-½) erf(√2 x)
, indeed satisfies the equation
and that together with
f(x) = e^(x²+½)
forms 2 linearly independent solutions (their Wronskian is non-zero as mentioned prior).
The product with the error function is no longer then dependent on = e^(x²+½)
Their Wronskian is identically equal to 2/√π which is not 0 so the two functions are 2 linearly independent solutions.
The third constant is set to 0 since y"/y must cancel and if y has a constant at the end then it won't cancel.
answer:
y = C₁ e^(x²+½)+√(π/2) C₂ e^(x²-½) erf(√2 x)
Z''+(z')^2=4x^2+2 WA gives the solution z=ln(c1+sqrt(2pi)erf(sqrt2 x))+c2+x^2 so the solution for your DE is : y=c2 e^(x^2)(c1+sqrt(2pi)erf(sqrt2x)). But don't ask me how 😂
Here’s what I think Wolfram Alpha’s solution is (letting y=f(x)):
f’’(x) =(4x^2+2)*f(x) -> f’’(x) -(4x^2+2)*f(x)=0. If you need replace x with x/2, you get f’’(x/2)-(x^2+2)*f(x/2). Now, let g(x)=f(x/2) -> g’’(x)=1/4*f’’(x/2). This means that 4g’’(x)-(x^2+2)*g(x)=0 -> g’’(x)-(1/2+x^2/4)*g(x)=0 -> g’’(x)+(-1+1/2-x^2/4)*g(x)=0. For whatever reason, one of the solutions to the more general equation y’’+(v+1/2-x^2/4)*y=0 is defined to be D_v(x), a parabolic cylinder function (also known as a Weber function). Because this is a second order linear differential equation, it has at most two linearly independent solutions. In this case, the other one is also a parabolic cylinder function: y=D_u(ix), where u=-1-v. For our specific example, g(x) must be a linear combination D_(-1)(x) and D_(0)(ix), so the original function f(x)=a*D_(-1)(2x)+b*D_0(2ix).
There is another method which is much simpler I would like to send to you if you don’t mind it is too simple and too short