Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 10480 by ssparasmani@gmail.com last updated on 13/Feb/17

d/dx(cos^2 xdy/dx)=0

$${d}/{dx}\left(\mathrm{cos}\:^{\mathrm{2}} {xdy}/{dx}\right)=\mathrm{0} \\ $$$$ \\ $$

Answered by mrW1 last updated on 13/Feb/17

d/dx(cos^2 xdy/dx)=0 ⇒cos^2 xdy/dx=C_1 dy/dx=(C_1 /(cos^2 x)) ⇒y=∫(C_1 /(cos^2 x))dx=C_1 tan x+C_2

$${d}/{dx}\left(\mathrm{cos}\:^{\mathrm{2}} {xdy}/{dx}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:^{\mathrm{2}} {xdy}/{dx}={C}_{\mathrm{1}} \\ $$$${dy}/{dx}=\frac{{C}_{\mathrm{1}} }{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$\Rightarrow{y}=\int\frac{{C}_{\mathrm{1}} }{\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}={C}_{\mathrm{1}} \mathrm{tan}\:{x}+{C}_{\mathrm{2}} \\ $$

Answered by robocop last updated on 14/Feb/17

((cos^2 x)dy)/dx ((1−sen^2 x)dy/dx dy/dx−(sen^2 x)dy/dx 1−2senxcosx d(1)/dx−d(2senxcosx)/dx=0 −2(senx(−senx)+cos^2 x=0 2sen^2 x−2cos^2 x=0 2−4cos^2 x=0 2cos^2 x=1 cosx=0.71 x=0.78

$$\left(\left({cos}^{\mathrm{2}} {x}\right){dy}\right)/{dx} \\ $$$$\left(\left(\mathrm{1}−{sen}^{\mathrm{2}} {x}\right){dy}/{dx}\right. \\ $$$${dy}/{dx}−\left({sen}^{\mathrm{2}} {x}\right){dy}/{dx} \\ $$$$\mathrm{1}−\mathrm{2}{senxcosx} \\ $$$$ \\ $$$${d}\left(\mathrm{1}\right)/{dx}−{d}\left(\mathrm{2}{senxcosx}\right)/{dx}=\mathrm{0} \\ $$$$−\mathrm{2}\left({senx}\left(−{senx}\right)+{cos}^{\mathrm{2}} {x}=\mathrm{0}\right. \\ $$$$\mathrm{2}{sen}^{\mathrm{2}} {x}−\mathrm{2}{cos}^{\mathrm{2}} {x}=\mathrm{0} \\ $$$$\mathrm{2}−\mathrm{4}{cos}^{\mathrm{2}} {x}=\mathrm{0} \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} {x}=\mathrm{1} \\ $$$${cosx}=\mathrm{0}.\mathrm{71} \\ $$$${x}=\mathrm{0}.\mathrm{78} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com