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Question Number 67860 Answers: 1 Comments: 3

Question Number 67852 Answers: 1 Comments: 4

Question Number 67851 Answers: 0 Comments: 5

find ∫ (dx/(x^2 −z)) with z from C .

$${find}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{z}}\:\:{with}\:{z}\:{from}\:{C}\:. \\ $$

Question Number 67850 Answers: 0 Comments: 1

calculate ∫_(−∞) ^(+∞) (dx/(x^2 −z)) with z from C

$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} −{z}}\:\:{with}\:{z}\:{from}\:{C} \\ $$

Question Number 67849 Answers: 1 Comments: 7

Question Number 67844 Answers: 0 Comments: 2

x^2 +∣x∣−6=0

$${x}^{\mathrm{2}} +\mid{x}\mid−\mathrm{6}=\mathrm{0} \\ $$

Question Number 67836 Answers: 0 Comments: 0

If determinant ((x^n ,x^(n+2) ,x^(n+3) ),(y^n ,y^(n+2) ,y^(n+3) ),(z^n ,z^(n+2) ,z^(n+3) )) = (x−y)(y−z)(z−x)((1/x) + (1/y) + (1/z)), then n equals

$$\mathrm{If}\:\begin{vmatrix}{{x}^{{n}} }&{{x}^{{n}+\mathrm{2}} }&{{x}^{{n}+\mathrm{3}} }\\{{y}^{{n}} }&{{y}^{{n}+\mathrm{2}} }&{{y}^{{n}+\mathrm{3}} }\\{{z}^{{n}} }&{{z}^{{n}+\mathrm{2}} }&{{z}^{{n}+\mathrm{3}} }\end{vmatrix} \\ $$$$=\:\left({x}−{y}\right)\left({y}−{z}\right)\left({z}−{x}\right)\left(\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}}\right), \\ $$$$\mathrm{then}\:{n}\:\mathrm{equals} \\ $$

Question Number 67835 Answers: 1 Comments: 0

∫_0 ^2 x(8−x^3 )^(1/3) dx

$$\int_{\mathrm{0}} ^{\mathrm{2}} {x}\left(\mathrm{8}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx} \\ $$

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)) :}

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

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