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Question Number 67826    Answers: 2   Comments: 4

Question Number 67823    Answers: 0   Comments: 0

∫_0 ^1 x^(lnx+e^(lnx/x) ) dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{lnx}+{e}^{{lnx}/{x}} } {dx} \\ $$

Question Number 67820    Answers: 0   Comments: 1

x^3 −x^2 −6x

$${x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{6}{x} \\ $$

Question Number 67819    Answers: 0   Comments: 1

y=x^5 +ax^4 +bx^3 +cx^2 +dx+e If we let x=t+h can we find h in terms of a,b,c,d,e such that y=(t+R)(t^2 +pt+q)(t^2 +s) this means two roots are of opposite sign, of course its possible by shifting the curve along x, but can we find the shift h ?

$${y}={x}^{\mathrm{5}} +{ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{dx}+{e} \\ $$$${If}\:{we}\:{let}\:{x}={t}+{h} \\ $$$${can}\:{we}\:{find}\:{h}\:{in}\:{terms}\:{of}\:{a},{b},{c},{d},{e} \\ $$$${such}\:{that} \\ $$$${y}=\left({t}+{R}\right)\left({t}^{\mathrm{2}} +{pt}+{q}\right)\left({t}^{\mathrm{2}} +{s}\right) \\ $$$${this}\:{means}\:{two}\:{roots}\:{are}\:{of} \\ $$$${opposite}\:{sign},\:{of}\:{course}\:{its} \\ $$$${possible}\:{by}\:{shifting}\:{the}\:{curve} \\ $$$${along}\:{x},\:{but}\:{can}\:{we}\:{find}\:{the} \\ $$$${shift}\:\boldsymbol{{h}}\:? \\ $$

Question Number 67807    Answers: 1   Comments: 2

Question Number 67799    Answers: 1   Comments: 3

calculate ∫_0 ^∞ (dx/((x^2 −e^(ia) )(x^2 −e^(ib) ))) with a>0 andb>0

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:−{e}^{{ia}} \right)\left({x}^{\mathrm{2}} −{e}^{{ib}} \right)}\:\:{with}\:{a}>\mathrm{0}\:{andb}>\mathrm{0} \\ $$

Question Number 67795    Answers: 0   Comments: 2

let A_p =∫_0 ^π x^p cos(nx)dx 1) calculate A_0 ,A_1 ,A_2 2)determine a relation of recurrence between A_p

$${let}\:\:{A}_{{p}} =\int_{\mathrm{0}} ^{\pi} \:{x}^{{p}} \:{cos}\left({nx}\right){dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{\mathrm{0}} ,{A}_{\mathrm{1}} ,{A}_{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){determine}\:{a}\:{relation}\:{of}\:{recurrence}\:{between}\:\:{A}_{{p}} \\ $$

Question Number 67775    Answers: 2   Comments: 0

Question Number 67774    Answers: 1   Comments: 0

Question Number 67772    Answers: 2   Comments: 3

Question Number 67770    Answers: 1   Comments: 0

Question Number 67813    Answers: 1   Comments: 8

Question Number 67760    Answers: 0   Comments: 0

y′=y^2 +2 ; y(0)=2 solve by picards iteration method

$${y}'={y}^{\mathrm{2}} +\mathrm{2}\:;\:{y}\left(\mathrm{0}\right)=\mathrm{2} \\ $$$$ \\ $$$${solve}\:{by}\:{picards}\:{iteration}\:{method} \\ $$$$ \\ $$

Question Number 67759    Answers: 0   Comments: 1

using variation of parameters method (x+2)^2 y′′−(x+2)y′=2x+4 x^2 y′′+2xy′−2y=x^2 lnx+3x

$${using}\:{variation}\:{of}\:{parameters}\:{method} \\ $$$$ \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} {y}''−\left({x}+\mathrm{2}\right){y}'=\mathrm{2}{x}+\mathrm{4} \\ $$$$ \\ $$$$ \\ $$$${x}^{\mathrm{2}} {y}''+\mathrm{2}{xy}'−\mathrm{2}{y}={x}^{\mathrm{2}} {lnx}+\mathrm{3}{x} \\ $$

Question Number 67761    Answers: 0   Comments: 0

solve by laplace transform method x^(•• ) +w_0 ^2 x=coswt x(0)=x_0 x^• (0)=v_0 w^2 ≠ w_0 ^2

$${solve}\:{by}\:{laplace}\:{transform}\:{method} \\ $$$$ \\ $$$$\overset{\bullet\bullet\:} {{x}}\:+{w}_{\mathrm{0}} ^{\mathrm{2}} {x}={coswt}\:\:\:{x}\left(\mathrm{0}\right)={x}_{\mathrm{0}} \:\overset{\bullet} {{x}}\left(\mathrm{0}\right)={v}_{\mathrm{0}} \:\:\:\:\:{w}^{\mathrm{2}} \neq\:{w}_{\mathrm{0}} ^{\mathrm{2}} \\ $$

Question Number 67762    Answers: 0   Comments: 0

solve by the complex method y^(iv) +3y^3 +2y^2 =−3sin2x y^2 +a^2 y=e^x cosecx

$${solve}\:{by}\:{the}\:{complex}\:{method} \\ $$$$ \\ $$$$ \\ $$$${y}^{{iv}} +\mathrm{3}{y}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} =−\mathrm{3}{sin}\mathrm{2}{x} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${y}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}={e}^{{x}} {cosecx} \\ $$

Question Number 67749    Answers: 1   Comments: 0

Question Number 67745    Answers: 3   Comments: 0

solve the system of equations { ((3∣x−5∣+4=y)),((∣y−3∣=4x−12)) :}

$${solve}\:{the}\:{system}\:{of}\:{equations\begin{cases}{\mathrm{3}\mid{x}−\mathrm{5}\mid+\mathrm{4}={y}}\\{\mid{y}−\mathrm{3}\mid=\mathrm{4}{x}−\mathrm{12}}\end{cases}} \\ $$

Question Number 67744    Answers: 0   Comments: 4

let f(x) =∫_0 ^∞ ((sin(t^2 ))/((x^2 +t^2 )^2 ))dt with x>0 1)determine a explicit form for f(x) 2) find also g(x) =∫_0 ^∞ ((sin(t^2 ))/((x^2 +t^2 )^3 ))dt 3) give f^((n)) (x) at form of integral and calculate f^((n)) (1). 4) find the valueof ∫_0 ^∞ ((sin(t^2 ))/((1+t^2 )^2 )) dt and ∫_0 ^∞ ((sin(t^2 ))/((1+t^2 )^3 ))dt

$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right){determine}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:{also}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{give}\:{f}^{\left({n}\right)} \left({x}\right)\:{at}\:{form}\:{of}\:{integral}\:{and}\:{calculate}\:{f}^{\left({n}\right)} \left(\mathrm{1}\right). \\ $$$$\left.\mathrm{4}\right)\:{find}\:{the}\:{valueof}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt} \\ $$

Question Number 67743    Answers: 0   Comments: 0

^

$$\:^{} \\ $$

Question Number 67728    Answers: 1   Comments: 5

Question Number 67719    Answers: 1   Comments: 0

Question Number 67716    Answers: 2   Comments: 0

Question Number 67711    Answers: 0   Comments: 0

Find the value of (1/(cos^2 (10°))) + (1/(sin^2 (20°))) + (1/(sin^2 (40°)))

$${Find}\:\:{the}\:\:{value}\:\:{of} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{10}°\right)}\:+\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{20}°\right)}\:+\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{40}°\right)}\: \\ $$

Question Number 67708    Answers: 1   Comments: 0

Question Number 67698    Answers: 0   Comments: 0

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