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- How to arrange $e^3,3^e,e^ {\pi},\pi^e,3^ {\pi},\pi^3$ in the . . .
This question shows research effort; it is useful and clear
- Solving $\int_0^ {\infty}x^3e^ {-x^2}dx$ [duplicate]
You used a substitution, and then integration by parts once But what if you start out with integration by parts? $$ \begin {align} \int x^3e^ {-x^2}\,dx =\int -\frac12x^2\left (-2xe^ {-x^2}\right)\,dx\\ =-\frac12x^2e^ {-x^2}+\int xe^ {-x^2}\,dx\\ =-\frac12x^2e^ {-x^2}-\frac12e^ {-x^2}+C \end {align} $$ And "from $0$ to $\infty$ " gives
- calculus - Integration by Parts Question: Integrate $x^3e^x . . .
A nice and quick way to visualize integration by parts (it could be a time-saver!): $$\matrix { \text {differentiate} \text {integrate} \\ x^3 e^x
- calculus - (Laplace Method) $y - 4y = 6e^ {3t} - 3e^ {-t . . .
Taking the Laplace Transform to both sides of the ODE we get \begin {align*} s^2Y (s)-s (1)- (-1)-4 [sY (s)-1] =\frac 6 {s-3}-\frac 3 {s+1}\\ [4pt] (s^2-4s)Y (s
- Find the general solution to $xy = 2y + x^3e^x$
Find the general solution to $xy' = 2y + x^3e^x$ Ask Question Asked 4 years ago Modified 4 years ago
- Contour Integration Integration by Parts: $\int _0^ {2\pi}\sin^2 . . .
I need to find the value of $\displaystyle \int _0^ {2\pi}\sin^2 \left (\frac {-\pi} {6}+3e^ {it} \right)dt$ I figured I could use contour integration and the Cauchy-Goursat theorem to do so
- If the Wronskian W of $f$ and $g$ is $3e^{4t}$, and if $f(t) = e^{2t . . .
If you simplify by factoring out $e^ {2t}$ and cancelling, that would give you $$g' - 2g = 3e^ {2t},$$ instead of $g'-2g = 3e^ {4t}$, which is what you had So it looks like you made a simplification error
- integration - Evaluation of $\int_ {0}^ {\infty}t^3e^ {-3t}dt . . .
Evaluation of $\int_ {0}^ {\infty}t^3e^ {-3t}dt$ Ask Question Asked 9 years, 10 months ago Modified 9 years, 10 months ago
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