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AllQuestion and Answers: Page 1187

Question Number 96231 Answers: 0 Comments: 1

Question Number 96222 Answers: 1 Comments: 2

If for nonzero x ; 2f (x^2 )+3f ((1/x^2 )) = x^2 −1 then f (x^2 ) = ?

$$\mathrm{If}\:\mathrm{for}\:\mathrm{nonzero}\:{x}\:;\:\mathrm{2}{f}\:\left({x}^{\mathrm{2}} \right)+\mathrm{3}{f}\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:=\:{x}^{\mathrm{2}} −\mathrm{1} \\ $$$${then}\:{f}\:\left({x}^{\mathrm{2}} \right)\:=\:? \\ $$

Question Number 96220 Answers: 0 Comments: 0

∫((tan(x))/x)dx ∫x tan(x) dx

$$\int\frac{{tan}\left({x}\right)}{{x}}{dx} \\ $$$$\int{x}\:{tan}\left({x}\right)\:{dx} \\ $$

Question Number 96217 Answers: 1 Comments: 0

(dy/dx) = (((y^2 −x^2 +y)/x))

$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\left(\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} +\mathrm{y}}{\mathrm{x}}\right)\: \\ $$

Question Number 96211 Answers: 3 Comments: 0

solve inside C (x−(1/x))^3 +(x−(1/x))^2 +(x−(1/x))+1 =0

$$\mathrm{solve}\:\mathrm{inside}\:\mathrm{C}\:\:\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{3}} \:+\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} \:+\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{1}\:=\mathrm{0} \\ $$

Question Number 96497 Answers: 2 Comments: 0

Question Number 96200 Answers: 2 Comments: 0

solve y^(′′) +y^′ −2y =xcosx with y^((2)) (0)=1 and y^′ (0) =−2

$$\mathrm{solve}\:\:\mathrm{y}^{''} \:+\mathrm{y}^{'} \:−\mathrm{2y}\:=\mathrm{xcosx}\:\:\mathrm{with}\:\mathrm{y}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\mathrm{1}\:\mathrm{and}\:\mathrm{y}^{'} \left(\mathrm{0}\right)\:=−\mathrm{2} \\ $$

Question Number 96198 Answers: 1 Comments: 0

calculate f(a) =∫_0 ^∞ ((cos(sh(2x)))/(x^2 +a^2 ))dx and g(a) =∫_0 ^∞ ((cos(sh(2x)))/((x^2 +a^2 )^2 )) (a>0)

$$\mathrm{calculate}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{sh}\left(\mathrm{2x}\right)\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} }\mathrm{dx}\:\mathrm{and}\:\mathrm{g}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{sh}\left(\mathrm{2x}\right)\right)}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:\:\left(\mathrm{a}>\mathrm{0}\right) \\ $$

Question Number 96197 Answers: 1 Comments: 0

calculate ∫_0 ^∞ ((ch(cosx−sinx))/(x^2 +4))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{ch}\left(\mathrm{cosx}−\mathrm{sinx}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}}\mathrm{dx} \\ $$

Question Number 96196 Answers: 1 Comments: 0

let g(x) =ln(sinx) developp g at fourier serie

$$\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)\:=\mathrm{ln}\left(\mathrm{sinx}\right)\:\:\mathrm{developp}\:\mathrm{g}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$

Question Number 96195 Answers: 1 Comments: 0

calculate ∫_0 ^∞ (arctan((1/x)))^2 dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\right)^{\mathrm{2}} \:\mathrm{dx} \\ $$

Question Number 96194 Answers: 2 Comments: 0

let f(x) =ln(cosx) developp f at fourier serie

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{ln}\left(\mathrm{cosx}\right)\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$

Question Number 96193 Answers: 0 Comments: 0

calculate ∫_0 ^(π/2) (ln(cosx))^2 dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{ln}\left(\mathrm{cosx}\right)\right)^{\mathrm{2}} \:\mathrm{dx} \\ $$

Question Number 96192 Answers: 0 Comments: 2

find a particular solution to the equation y′ =(y/x)+sin(y/x) with original condition y(1)=(π/2)

$$\mathrm{find}\:\mathrm{a}\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{y}'\:=\frac{\mathrm{y}}{\mathrm{x}}+\mathrm{sin}\frac{\mathrm{y}}{\mathrm{x}}\:\mathrm{with}\:\mathrm{original}\:\mathrm{condition} \\ $$$$\mathrm{y}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}} \\ $$

Question Number 96189 Answers: 3 Comments: 0

find a common roots from the two quadratic eq 24x^2 +(p+4)x−1=0 and 6x^2 +11x+p+2=0

$$\mathrm{find}\:\mathrm{a}\:\mathrm{common}\:\mathrm{roots}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{quadratic}\:\mathrm{eq} \\ $$$$\mathrm{24x}^{\mathrm{2}} +\left(\mathrm{p}+\mathrm{4}\right)\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{6x}^{\mathrm{2}} +\mathrm{11x}+\mathrm{p}+\mathrm{2}=\mathrm{0} \\ $$

Question Number 96185 Answers: 1 Comments: 0

what are critical points of this function z = xy+5xy^2 +10y

$$\mathrm{what}\:\mathrm{are}\:\mathrm{critical}\:\mathrm{points}\:\mathrm{of}\:\mathrm{this} \\ $$$$\mathrm{function}\:\mathrm{z}\:=\:\mathrm{xy}+\mathrm{5xy}^{\mathrm{2}} +\mathrm{10y} \\ $$

Question Number 96182 Answers: 1 Comments: 0

x^2 y′′−xy′+y = 0

$${x}^{\mathrm{2}} {y}''−{xy}'+\mathrm{y}\:=\:\mathrm{0}\: \\ $$

Question Number 96175 Answers: 1 Comments: 1

∫(dx/(√(4x^2 +4x+3)))=?

$$\int\frac{{dx}}{\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}}=? \\ $$

Question Number 96171 Answers: 0 Comments: 2

(4+(√(15)))^(3/2) −(4−(√(15)))^(3/2) = k(√6) find k

$$\left(\mathrm{4}+\sqrt{\mathrm{15}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\:\mathrm{k}\sqrt{\mathrm{6}} \\ $$$$\mathrm{find}\:\mathrm{k}\: \\ $$

Question Number 96161 Answers: 1 Comments: 1

∫_1 ^4 ((sech^2 ((√x))+tanh ((√x)))/((√x) )) dx ?

$$\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:\frac{\mathrm{sech}\:^{\mathrm{2}} \left(\sqrt{{x}}\right)+\mathrm{tanh}\:\left(\sqrt{{x}}\right)}{\sqrt{{x}}\:}\:{dx}\:? \\ $$

Question Number 96155 Answers: 1 Comments: 4

Question Number 96148 Answers: 2 Comments: 14

Question Number 96140 Answers: 1 Comments: 5

xy′+y^2 =x^2 e^x ⇒ y′=xe^x −(y^2 /x) ⇒ y=xye^x −(1/(3x))∙y^3 ⇒ (y^3 /(3x))+y−xye^x =0 y((y^2 /(3x))+1−xe^x )=0 ⇒ y=±(√(3x(xe^x −1)))

$$\boldsymbol{{xy}}'+\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{e}}^{\boldsymbol{{x}}} \:\Rightarrow\:\boldsymbol{{y}}'=\boldsymbol{{xe}}^{\boldsymbol{{x}}} −\frac{\boldsymbol{{y}}^{\mathrm{2}} }{\boldsymbol{{x}}}\:\Rightarrow \\ $$$$\boldsymbol{{y}}=\boldsymbol{{xye}}^{\boldsymbol{{x}}} −\frac{\mathrm{1}}{\mathrm{3}\boldsymbol{{x}}}\centerdot\boldsymbol{{y}}^{\mathrm{3}} \Rightarrow\:\frac{\boldsymbol{{y}}^{\mathrm{3}} }{\mathrm{3}\boldsymbol{{x}}}+\boldsymbol{{y}}−\boldsymbol{{xye}}^{\boldsymbol{{x}}} =\mathrm{0} \\ $$$$\boldsymbol{{y}}\left(\frac{\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{3}\boldsymbol{{x}}}+\mathrm{1}−\boldsymbol{{xe}}^{\boldsymbol{{x}}} \right)=\mathrm{0}\:\Rightarrow\:\boldsymbol{{y}}=\pm\sqrt{\mathrm{3}\boldsymbol{{x}}\left(\boldsymbol{{xe}}^{\boldsymbol{{x}}} −\mathrm{1}\right)} \\ $$

Question Number 96138 Answers: 0 Comments: 1

xy′ + y^2 = x^2 e^x

$${xy}'\:+\:{y}^{\mathrm{2}} \:=\:{x}^{\mathrm{2}} {e}^{{x}} \: \\ $$

Question Number 96135 Answers: 0 Comments: 5

Question Number 96128 Answers: 1 Comments: 0

find ∫∫_R (x+2y)^2 dxdy in R=[−1,2] ×[0,2]

$${find}\:\int\int_{{R}} \:\left({x}+\mathrm{2}{y}\right)^{\mathrm{2}} \:{dxdy}\:{in}\:{R}=\left[−\mathrm{1},\mathrm{2}\right]\:×\left[\mathrm{0},\mathrm{2}\right]\: \\ $$

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