# Tag Info

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### what more can I say using Sturm's comparison theorem?

Sturm's comparison theorem is not the correct approach. If we set up $x(t) = \sqrt{x}u(\frac{x^2}{2})$, then $u$ satisfies Bessel equation. Using the asymptotic decay of Bessel function we can ...
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### How does one do this translation?

I have found the original paper you've taken the screenshot from and here's the deal. Your approach is absolutely correct, you shouldn't be confused with the constant terms because later on they are ...

### Diferencial Equations solve $\frac{e^{2x}+2x}{e^{x}}dy-(1+2y)dx=0$

This function can't be integrated .Only a part of this equation can be integrated and not the full .Note that many functions like this can't be integrated or anti derivative can't be found .This can ...

### Asymptotic Expansion of Integrals via Integration by Parts - Leading Behavior

$$\int_1^\infty \frac{\cos(xt)}{t}\,dt=-\text{Ci}(x)$$ Expanded as series $$\int_1^\infty \frac{\cos(xt)}{t}\,dt=-\log (x)-\gamma+\sum _{n=1}^\infty (-1)^{n+1} \frac {x^{2n}}{ 2n\,(2n)! }$$
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### Diferencial Equations solve $\frac{e^{2x}+2x}{e^{x}}dy-(1+2y)dx=0$

I am afraid that only a series solution would give a result. For example (using @Travis Willse's suggestion in comments), this would give $$y=C +(2C+1)\sum_{n=1}^\infty (-1)^{n+1}\, a_n\, x^n$$ the ...
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### Is there a formula for the Butcher tableau for the implicit Runge Kutta method for an arbitrary number of stages?

Yes ! Well, more or less. You'll find this in the following book for collocation methods (like Gauss). Hairer, Ernst; Lubich, Christian; Wanner, Gerhard, Geometric numerical integration. Structure-...
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### Why is the uniqueness theorem for 1st order differential equations involve taking a partial dervative?

As peE_ pointed out Picard-Lindelöf theorem requires Lipschitz continuity for the unique existence of the solution. If the function is not Lipschitz continuous then such existence is not guaranteed. ...
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### Why is the uniqueness theorem for 1st order differential equations involve taking a partial dervative?

Let $f:\Omega \subset \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$, $x'(t) = f(t,x)$ with initial condition $(t_{0},x_{0})$. For Picard-Lindelöf theorem it is needed $f(t,x)$ to be ...
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### How do I check if set of solutions to differential equation form vector subspace

You can apply the usual characterisation of subspaces: Check that $0$ is a solution to the equation; then take any solutions $x(t), y(t)$ (no need to find them explicitly, work formally) and verify ...
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### How do I check if set of solutions to differential equation form vector subspace

Like with any subspace question you have to make sure , a)zero is in it, with f, r*f is in it and with f and g f+g are in it.
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### Recovering a linear ODE w/ constant coefficients given the roots of the auxiliary equation

Hint: Hint 2: Sketch of solution:
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### Proving $\int\limits_{Q} u^2 dx ≤ C(n) \int\limits_{Q} (u^2_{x1} + · · · + u^2_{xn}) dx$ over a cube $Q$

One possible proof would be via developing $u$ into its Fourier series and then using simple formula for second norm (so called Parselval's theorem). In case of $n=1$, say $u = \sum a_ke^{2{\pi}kx}$, ...
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### An differential equation question that seems exact but not

We rearrange the terms to render $[(x)dy+(y)dx]+[x dx]-[y dy] = 0.$ Then the first term in brackets is just tge differential of $xy$ by the Product Rule, and the other terms are one-variable ...
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### An differential equation question that seems exact but not

You are correct to think that is an exact DE, the only obstacle on your way is that the total differential has undergone arithmetic transformations. To see this, let us solve the second equation and ...
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### Solve $\frac{d}{dx}f(x)=f(x-1)$

Using the Laplace transform $$s \hat f(s) = e^{-s}\hat f(s) + e^{-s}\int_{-1}^0e^{-sx}f(x)dx+f_0$$ Here $f(x),x\in[-1,0)$ is considered known, kind of initial conditions. Then the Laplace transform ...
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Note that $$\mathcal L \{te^{-3t}\}=\left.\mathcal{L}\{t\}\right|_{s\to s+3}=\frac{1}{(s+3)^2}$$ So instead we get:  \begin{align} s^2Y(s)+7sY(s)+10Y(s) = \frac{4}{(s+3)^2}-1\implies y(x) = \mathcal ...