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Question Number 169526 by ali009 last updated on 01/May/22
solve the D.E.  y^′ =tan(x+y)−1
$${solve}\:{the}\:{D}.{E}. \\ $$$${y}^{'} ={tan}\left({x}+{y}\right)−\mathrm{1} \\ $$
Answered by mr W last updated on 01/May/22
let u=x+y  (du/dx)=1+(dy/dx)  (du/dx)−1=tan u−1  (du/dx)=tan u  ((cos u)/(sin u))du=dx  ∫((cos u)/(sin u))du=∫dx  ∫(1/(sin u))d(sin u)=∫dx  ln (sin u)=x+C  sin u=Ce^x   u=sin^(−1) (Ce^x )  x+y=sin^(−1) (Ce^x )  ⇒y=sin^(−1) (Ce^x )−x
$${let}\:{u}={x}+{y} \\ $$$$\frac{{du}}{{dx}}=\mathrm{1}+\frac{{dy}}{{dx}} \\ $$$$\frac{{du}}{{dx}}−\mathrm{1}=\mathrm{tan}\:{u}−\mathrm{1} \\ $$$$\frac{{du}}{{dx}}=\mathrm{tan}\:{u} \\ $$$$\frac{\mathrm{cos}\:{u}}{\mathrm{sin}\:{u}}{du}={dx} \\ $$$$\int\frac{\mathrm{cos}\:{u}}{\mathrm{sin}\:{u}}{du}=\int{dx} \\ $$$$\int\frac{\mathrm{1}}{\mathrm{sin}\:{u}}{d}\left(\mathrm{sin}\:{u}\right)=\int{dx} \\ $$$$\mathrm{ln}\:\left(\mathrm{sin}\:{u}\right)={x}+{C} \\ $$$$\mathrm{sin}\:{u}={Ce}^{{x}} \\ $$$${u}=\mathrm{sin}^{−\mathrm{1}} \left({Ce}^{{x}} \right) \\ $$$${x}+{y}=\mathrm{sin}^{−\mathrm{1}} \left({Ce}^{{x}} \right) \\ $$$$\Rightarrow{y}=\mathrm{sin}^{−\mathrm{1}} \left({Ce}^{{x}} \right)−{x} \\ $$
Commented by ali009 last updated on 01/May/22
thank you
$${thank}\:{you}\: \\ $$
Commented by peter frank last updated on 02/May/22
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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