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Question Number 219798 Answers: 0 Comments: 0

solve y^((2)) (t)=y^((1)) (t)e^(−y(t))

$$\mathrm{solve}\: \\ $$$${y}^{\left(\mathrm{2}\right)} \left({t}\right)={y}^{\left(\mathrm{1}\right)} \left({t}\right){e}^{−{y}\left({t}\right)} \\ $$

Question Number 219797 Answers: 0 Comments: 0

solve (y^((2)) (t))^2 =(1/(1+y(t)))

$$\mathrm{solve} \\ $$$$\left({y}^{\left(\mathrm{2}\right)} \left({t}\right)\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}+{y}\left({t}\right)} \\ $$

Question Number 219796 Answers: 0 Comments: 0

Solve y^((2)) (t)=(y(t))^2 −ay^((1)) (t)−by(t)

$$\mathrm{Solve} \\ $$$${y}^{\left(\mathrm{2}\right)} \left({t}\right)=\left({y}\left({t}\right)\right)^{\mathrm{2}} −{ay}^{\left(\mathrm{1}\right)} \left({t}\right)−{by}\left({t}\right) \\ $$

Question Number 219795 Answers: 1 Comments: 0

∫_(D=[0,1]^N ) Π_(h=1) ^N e^(−(1/2)x_h ) dx_h

$$\int_{\mathcal{D}=\left[\mathrm{0},\mathrm{1}\right]^{{N}} } \:\underset{{h}=\mathrm{1}} {\overset{{N}} {\prod}}\:{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{x}_{{h}} } \mathrm{d}{x}_{{h}} \\ $$

Question Number 219794 Answers: 0 Comments: 0

∫_0 ^( ∞) ((Y_0 (t)e^(−3t) )/(t^2 +1)) dt

$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{Y}_{\mathrm{0}} \left({t}\right){e}^{−\mathrm{3}{t}} }{{t}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{t} \\ $$

Question Number 219793 Answers: 1 Comments: 0

∫_0 ^( ∞) (e^(−t) /(t^2 +1)) dt=?

$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{e}^{−{t}} }{{t}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{t}=? \\ $$

Question Number 219792 Answers: 1 Comments: 0

lim_(z→∞) ((J_1 (z))/(Y_0 (z))) lim_(z→0) z((1−(1/2)z^2 −cos((z/(1−z^2 ))))/z^4 ) lim_(z→0) ((J_ν (z+h)Y_ν (z)−J_ν (z)Y_ν (z))/h)

$$\underset{{z}\rightarrow\infty} {\mathrm{lim}}\:\frac{{J}_{\mathrm{1}} \left({z}\right)}{{Y}_{\mathrm{0}} \left({z}\right)} \\ $$$$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{z}\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{z}^{\mathrm{2}} −\mathrm{cos}\left(\frac{{z}}{\mathrm{1}−{z}^{\mathrm{2}} }\right)}{{z}^{\mathrm{4}} } \\ $$$$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{J}_{\nu} \left({z}+{h}\right){Y}_{\nu} \left({z}\right)−{J}_{\nu} \left({z}\right){Y}_{\nu} \left({z}\right)}{{h}}\: \\ $$

Question Number 219789 Answers: 1 Comments: 4

Question Number 219787 Answers: 2 Comments: 0

Question Number 219786 Answers: 1 Comments: 0

Question Number 219785 Answers: 3 Comments: 0

Question Number 219784 Answers: 8 Comments: 0

Question Number 219871 Answers: 1 Comments: 0

F(s)=∫_0 ^( ∞) ((sin(t))/(t^2 +α^2 ))e^(−st) dt F(3)=??

$${F}\left({s}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{sin}\left({t}\right)}{{t}^{\mathrm{2}} +\alpha^{\mathrm{2}} }{e}^{−{st}} \:\mathrm{d}{t} \\ $$$$\mathrm{F}\left(\mathrm{3}\right)=?? \\ $$

Question Number 219770 Answers: 1 Comments: 3

Question Number 219769 Answers: 2 Comments: 1

Question Number 219763 Answers: 3 Comments: 0

Question Number 219761 Answers: 2 Comments: 0

Question Number 219733 Answers: 2 Comments: 0

Question Number 219732 Answers: 1 Comments: 0

Question Number 219731 Answers: 1 Comments: 0

Question Number 219730 Answers: 0 Comments: 0

f(w)=∫_0 ^( ∞) ((θ(s−1))/(s(s−1)^α ))e^(−sw) ds θ^ (s)= { ((0 s<0)),((1 s>0)) :}

$${f}\left({w}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\frac{\theta\left({s}−\mathrm{1}\right)}{{s}\left({s}−\mathrm{1}\right)^{\alpha} }{e}^{−{sw}} \:\mathrm{d}{s} \\ $$$$\hat {\theta}\left({s}\right)=\begin{cases}{\mathrm{0}\:\:{s}<\mathrm{0}}\\{\mathrm{1}\:\:{s}>\mathrm{0}}\end{cases} \\ $$

Question Number 219725 Answers: 0 Comments: 1

Question Number 219724 Answers: 1 Comments: 0

prove; Π_(n=1) ^∞ (((2n+1)^3 −3(2n+1)+2)/((2n+1)^3 )) = (π/6)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}; \\ $$$$\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\:\frac{\pi}{\mathrm{6}} \\ $$$$ \\ $$

Question Number 219723 Answers: 2 Comments: 0

Question Number 219722 Answers: 2 Comments: 0

Question Number 219721 Answers: 5 Comments: 0

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