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AllQuestion and Answers: Page 1194

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]

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

Question Number 96131    Answers: 1   Comments: 0

(((x+4)^2 ))^(1/(3 )) + 4 (((x−3)^2 ))^(1/(3 )) + 5 ((x^2 +x−12))^(1/(3 )) = 0

$$\sqrt[{\mathrm{3}\:\:}]{\left({x}+\mathrm{4}\right)^{\mathrm{2}} }\:+\:\mathrm{4}\:\sqrt[{\mathrm{3}\:\:}]{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:+\:\mathrm{5}\:\sqrt[{\mathrm{3}\:\:}]{{x}^{\mathrm{2}} +{x}−\mathrm{12}}\:=\:\mathrm{0} \\ $$

Question Number 96125    Answers: 1   Comments: 0

4x^2 y′′ +12xy′ + 3y = 0

$$\mathrm{4x}^{\mathrm{2}} \mathrm{y}''\:+\mathrm{12xy}'\:+\:\mathrm{3y}\:=\:\mathrm{0} \\ $$

Question Number 96117    Answers: 2   Comments: 0

(1−2xy) dx + (4y^3 −x^2 ) dy = 0

$$\left(\mathrm{1}−\mathrm{2}{xy}\right)\:{dx}\:+\:\left(\mathrm{4}{y}^{\mathrm{3}} −{x}^{\mathrm{2}} \right)\:{dy}\:=\:\mathrm{0}\: \\ $$

Question Number 96108    Answers: 0   Comments: 0

are the system (z,+,≤)is orderd integral domain ?

$${are}\:{the}\:{system}\:\left({z},+,\leqslant\right){is}\:{orderd}\:{integral}\:{domain}\:? \\ $$

Question Number 96106    Answers: 0   Comments: 1

Question Number 96097    Answers: 1   Comments: 0

Question Number 96092    Answers: 0   Comments: 4

find ∫(dx/(tan^(−1) (x)))

$${find}\:\int\frac{{dx}}{{tan}^{−\mathrm{1}} \left({x}\right)} \\ $$

Question Number 97266    Answers: 2   Comments: 1

If x &y satisfy the equation x^2 +y^2 −4x−6y−1 =0 find minimum value of x+y ?

$$\mathrm{If}\:\mathrm{x}\:\&\mathrm{y}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{4x}−\mathrm{6y}−\mathrm{1}\:=\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}\:? \\ $$

Question Number 96089    Answers: 2   Comments: 0

Question Number 96083    Answers: 1   Comments: 0

∫ ((e^x (1+sin x))/(1+cos x)) dx

$$\int\:\frac{\mathrm{e}^{\mathrm{x}} \left(\mathrm{1}+\mathrm{sin}\:\mathrm{x}\right)}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$

Question Number 96078    Answers: 0   Comments: 0

lim_((x,y)→(1,2)) sin (((x−1)/(y−2)))

$$\underset{\left({x},\mathrm{y}\right)\rightarrow\left(\mathrm{1},\mathrm{2}\right)} {\mathrm{lim}}\:\mathrm{sin}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{y}−\mathrm{2}}\right)\: \\ $$

Question Number 96076    Answers: 2   Comments: 0

∫ 3x.2^x dx ?

$$\int\:\mathrm{3x}.\mathrm{2}^{\mathrm{x}} \:\mathrm{dx}\:?\: \\ $$

Question Number 96072    Answers: 1   Comments: 0

(x^2 +24x+24).(x^2 +x+24)= 24x^2

$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{24x}+\mathrm{24}\right).\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{24}\right)=\:\mathrm{24x}^{\mathrm{2}} \\ $$

Question Number 96065    Answers: 3   Comments: 0

y′ + y = x (y^2 )^(1/(3 ))

$$\mathrm{y}'\:+\:\mathrm{y}\:=\:\mathrm{x}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{y}^{\mathrm{2}} } \\ $$

Question Number 96114    Answers: 0   Comments: 6

A read only snapshot of questions and answers on this forum is accessible from www.tinkutara.com. This can be viewed in any browser and also included plain text. Version 2.079 has been uploaded to playstore and will be available for download in next couple of days.

$$\mathrm{A}\:\mathrm{read}\:\mathrm{only}\:\mathrm{snapshot}\:\mathrm{of}\:\mathrm{questions} \\ $$$$\mathrm{and}\:\mathrm{answers}\:\mathrm{on}\:\mathrm{this}\:\mathrm{forum}\:\mathrm{is} \\ $$$$\mathrm{accessible}\:\mathrm{from}\:\mathrm{www}.\mathrm{tinkutara}.\mathrm{com}. \\ $$$$\mathrm{This}\:\mathrm{can}\:\mathrm{be}\:\mathrm{viewed}\:\mathrm{in}\:\mathrm{any}\:\mathrm{browser} \\ $$$$\mathrm{and}\:\mathrm{also}\:\mathrm{included}\:\mathrm{plain}\:\mathrm{text}. \\ $$$$ \\ $$$$\mathrm{Version}\:\mathrm{2}.\mathrm{079}\:\mathrm{has}\:\mathrm{been}\:\mathrm{uploaded} \\ $$$$\mathrm{to}\:\mathrm{playstore}\:\mathrm{and}\:\mathrm{will}\:\mathrm{be}\:\mathrm{available} \\ $$$$\mathrm{for}\:\mathrm{download}\:\mathrm{in}\:\mathrm{next}\:\mathrm{couple}\:\mathrm{of}\:\mathrm{days}. \\ $$

Question Number 96060    Answers: 1   Comments: 4

Question Number 96052    Answers: 1   Comments: 0

Question Number 96051    Answers: 0   Comments: 1

(x−y) dx + (x^2 +y^2 ) dy = 0

$$\left(\mathrm{x}−\mathrm{y}\right)\:\mathrm{dx}\:+\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dy}\:=\:\mathrm{0}\: \\ $$

Question Number 96041    Answers: 1   Comments: 0

if f(x)=3^x^(2+x^3 ) find f′(x)?

$${if}\:{f}\left({x}\right)=\mathrm{3}^{{x}^{\mathrm{2}+{x}^{\mathrm{3}} } } \:\:\:{find}\:{f}'\left({x}\right)? \\ $$

Question Number 96034    Answers: 2   Comments: 1

∫_0 ^∞ (1/(x^(10) +1))dx=((2π)/(5((√5)−1)))=((πφ)/5)

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{10}} +\mathrm{1}}{dx}=\frac{\mathrm{2}\pi}{\mathrm{5}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}=\frac{\pi\phi}{\mathrm{5}} \\ $$

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