Question Number 212612 by hardmath last updated on 18/Oct/24
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}: \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:\:=\:\:\mathrm{cos}\left(\mathrm{x}\:+\:\mathrm{y}\right) \\ $$
Answered by mr W last updated on 19/Oct/24
$${let}\:{t}={x}+{y} \\ $$$$\frac{{dt}}{{dx}}=\mathrm{1}+\frac{{dy}}{{dx}} \\ $$$$\frac{{dt}}{{dx}}−\mathrm{1}=\mathrm{cos}\:{t} \\ $$$$\frac{{dt}}{\mathrm{cos}\:{t}+\mathrm{1}}={dx} \\ $$$$\int\frac{{dt}}{\mathrm{cos}\:{t}+\mathrm{1}}=\int{dx} \\ $$$$\int\frac{{dt}}{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{{t}}{\mathrm{2}}}=\int{dx} \\ $$$$\mathrm{tan}\:\frac{{t}}{\mathrm{2}}={x}+{C} \\ $$$${t}=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left({x}+{C}\right) \\ $$$${x}+{y}=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left({x}+{C}\right) \\ $$$$\Rightarrow{y}=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left({x}+{C}\right)−{x}\:\checkmark \\ $$
Commented by hardmath last updated on 19/Oct/24
$$\mathrm{thankyou}\:\mathrm{dearprofessor} \\ $$