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Feb 12 |
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PDE method of characteristics solving $e^{t^2}u_t+tu_x=0$ with $u(x,0)=x+2$ @user61933 I've edited the answer in order to address your questions. Please, if you find it appropriate, vote up and accept the answer. |
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Feb 12 |
revised |
PDE method of characteristics solving $e^{t^2}u_t+tu_x=0$ with $u(x,0)=x+2$ added 524 characters in body |
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Feb 12 |
answered | PDE method of characteristics solving $e^{t^2}u_t+tu_x=0$ with $u(x,0)=x+2$ |
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Feb 12 |
revised |
PDE method of characteristics solving $e^{t^2}u_t+tu_x=0$ with $u(x,0)=x+2$ texified it |
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Feb 12 |
reviewed | Reviewed Construct a convergent series of positive terms with $\displaystyle\limsup_{n\to\infty} {a_{n+1}\over{a_n}}=\infty$ |
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Feb 12 |
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PDE method of characteristics solving $e^{t^2}u_t+tu_x=0$ with $u(x,0)=x+2$ I added $\LaTeX$ to your question. Please verify it's correct. |
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Feb 12 |
suggested | suggested edit on PDE method of characteristics solving $e^{t^2}u_t+tu_x=0$ with $u(x,0)=x+2$ |
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Feb 12 |
revised |
Does the limit as $(x,y) \to (1,2)$ of $3x^3-x^2 y^2$ exist? texified it |
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Feb 12 |
suggested | suggested edit on Does the limit as $(x,y) \to (1,2)$ of $3x^3-x^2 y^2$ exist? |
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Jan 30 |
revised |
Integral using residue theorem (maybe) edited tags |
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Jan 30 |
answered | Integral using residue theorem (maybe) |
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Jan 30 |
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Diff eq. transformation polar coordinates Two things. Your notation is obscure, as $t$ is function of the variable of integration, i.e. $\frac{d t}{d \xi} \equiv {t}'$, hence you are using the chain rule wrong; also, the cubic term on the second component of your substitution should read $-r^3 \sin t$. See my answer for details. |
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Jan 30 |
answered | Diff eq. transformation polar coordinates |
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Jan 14 |
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How to find the following derivative? @LanceFerd It's the domain of definition for $\arcsin t$, i.e. the range where $\sin x$ is invertible -in the principal branch, that is-. |
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Jan 14 |
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How to find the following derivative? @LanceFerd One has to be carefull when taking root of a squared function: $$\sqrt{f^2(x)} = \big|f(x)\big|.$$ See my answer for details. |
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Jan 14 |
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How to find the following derivative? added 116 characters in body |
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Jan 14 |
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How to find the following derivative? @GitGud Edited. By the way, if you use \arcsin x, you produce $$\arcsin x.$$ |
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Jan 14 |
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How to find the following derivative? added 484 characters in body |
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Jan 14 |
answered | How to find the following derivative? |
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Jan 13 |
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Wave equation solution properties @rlgordonma I'd say that the fact that $v_{\alpha \beta} = 0$ implies that $v(\alpha,\beta) = F(\beta) + G(\alpha)$ which, in turn, means that $u(x,t) = F(x-t) + G(x+t)$ (this can be ensured because the transformation $(x,t) \to (\alpha,\beta)$ is invertible), and given that the wave equation with initial conditions has a unique solution (provided $f$ and $g$ are in some space, details left to the OP), this has to be it. Finally, using the initial conditions, one can determine $F$ and $G$ in terms of $f$ and $g$, and derive d'Alambert solution. |