Menu Close

dy-dx-sec-x-y-




Question Number 85503 by john santu last updated on 22/Mar/20
(dy/dx) = sec (x+y)
$$\frac{{dy}}{{dx}}\:=\:\mathrm{sec}\:\left({x}+{y}\right)\: \\ $$
Answered by mr W last updated on 22/Mar/20
u=x+y  (du/dx)=1+(dy/dx)  (du/dx)−1=(1/(cos u))  ((cos u du)/(1+cos u))=dx  ∫((cos u du)/(1+cos u))=∫dx  ∫(1−(1/(1+cos u)))du=∫dx  u−tan (u/2)=x+C  ⇒y−tan ((x+y)/2)=C
$${u}={x}+{y} \\ $$$$\frac{{du}}{{dx}}=\mathrm{1}+\frac{{dy}}{{dx}} \\ $$$$\frac{{du}}{{dx}}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{cos}\:{u}} \\ $$$$\frac{\mathrm{cos}\:{u}\:{du}}{\mathrm{1}+\mathrm{cos}\:{u}}={dx} \\ $$$$\int\frac{\mathrm{cos}\:{u}\:{du}}{\mathrm{1}+\mathrm{cos}\:{u}}=\int{dx} \\ $$$$\int\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:{u}}\right){du}=\int{dx} \\ $$$${u}−\mathrm{tan}\:\frac{{u}}{\mathrm{2}}={x}+{C} \\ $$$$\Rightarrow{y}−\mathrm{tan}\:\frac{{x}+{y}}{\mathrm{2}}={C} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *