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Question Number 95068 by Mr.D.N. last updated on 23/May/20

   Solve the differential equations−:     1. (dy/dx) = sin(x+y)+ cos(x+y)

$$\:\:\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equations}−: \\ $$$$\:\:\:\mathrm{1}.\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)+\:\mathrm{cos}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\: \\ $$

Commented by EmericGent last updated on 22/May/20

Would be easy with cos(x-y) instead of cos(x+y)

Commented by EmericGent last updated on 23/May/20

Do we have to use Sturm Liouville in order to solve the second? (I'm sorry I'm learning maths it's hard for me)

Commented by bobhans last updated on 23/May/20

(dy/dx) = sin (x+y)+sin ((π/2)−(x+y))  (dy/dx) = 2sin ((π/4)) cos (x+y−(π/4))  (dy/dx) = (√2) cos (x+y−(π/4))   set x+y−(π/4) = v ⇒1+(dy/dx) = (dv/dx)  (dv/dx)−1 = (√2) cos v ⇒(dv/dx) = 1+(√2) cos v  (dv/(1+(√2) cos v)) = dx ⇒∫ (dv/(1+(√2) cos v)) = x+c

$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{sin}\:\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\left(\mathrm{x}+\mathrm{y}\right)\right) \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{4}}\right)\:\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}−\frac{\pi}{\mathrm{4}}\right)\: \\ $$$$\mathrm{set}\:\mathrm{x}+\mathrm{y}−\frac{\pi}{\mathrm{4}}\:=\:{v}\:\Rightarrow\mathrm{1}+\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{{dv}}{{dx}} \\ $$$$\frac{{dv}}{{dx}}−\mathrm{1}\:=\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:{v}\:\Rightarrow\frac{{dv}}{{dx}}\:=\:\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{cos}\:{v} \\ $$$$\frac{{dv}}{\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{cos}\:{v}}\:=\:{dx}\:\Rightarrow\int\:\frac{{dv}}{\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{cos}\:{v}}\:=\:{x}+{c} \\ $$$$ \\ $$

Answered by mr W last updated on 23/May/20

1.  let u=x+y  (du/dx)=1+(dy/dx)  (du/dx)−1=sin u+cos u  (du/(1+sin u+cos u))=dx  ∫(du/(1+(√2)sin (u+(π/4))))=∫dx  ln(((tan ((u+(π/4))/2)+(√2)−1)/(tan ((u+(π/4))/2)+(√2)+1)))=x+C  ln(((tan ((x+y+(π/4))/2)+(√2)−1)/(tan ((x+y+(π/4))/2)+(√2)+1)))=x+C  ln(1−(2/(tan ((x+y+(π/4))/2)+(√2)+1)))=x+C  1−(2/(tan ((x+y+(π/4))/2)+(√2)+1))=Ce^x   tan ((x+y+(π/4))/2)=(2/(1−Ce^x ))−1−(√2)  ⇒y=2 tan^(−1) ((2/(1−Ce^x ))−1−(√2))−(π/4)−x

$$\mathrm{1}. \\ $$$${let}\:{u}={x}+{y} \\ $$$$\frac{{du}}{{dx}}=\mathrm{1}+\frac{{dy}}{{dx}} \\ $$$$\frac{{du}}{{dx}}−\mathrm{1}=\mathrm{sin}\:{u}+\mathrm{cos}\:{u} \\ $$$$\frac{{du}}{\mathrm{1}+\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}={dx} \\ $$$$\int\frac{{du}}{\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{sin}\:\left({u}+\frac{\pi}{\mathrm{4}}\right)}=\int{dx} \\ $$$$\mathrm{ln}\left(\frac{\mathrm{tan}\:\frac{{u}+\frac{\pi}{\mathrm{4}}}{\mathrm{2}}+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{tan}\:\frac{{u}+\frac{\pi}{\mathrm{4}}}{\mathrm{2}}+\sqrt{\mathrm{2}}+\mathrm{1}}\right)={x}+{C} \\ $$$$\mathrm{ln}\left(\frac{\mathrm{tan}\:\frac{{x}+{y}+\frac{\pi}{\mathrm{4}}}{\mathrm{2}}+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{tan}\:\frac{{x}+{y}+\frac{\pi}{\mathrm{4}}}{\mathrm{2}}+\sqrt{\mathrm{2}}+\mathrm{1}}\right)={x}+{C} \\ $$$$\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{tan}\:\frac{{x}+{y}+\frac{\pi}{\mathrm{4}}}{\mathrm{2}}+\sqrt{\mathrm{2}}+\mathrm{1}}\right)={x}+{C} \\ $$$$\mathrm{1}−\frac{\mathrm{2}}{\mathrm{tan}\:\frac{{x}+{y}+\frac{\pi}{\mathrm{4}}}{\mathrm{2}}+\sqrt{\mathrm{2}}+\mathrm{1}}={Ce}^{{x}} \\ $$$$\mathrm{tan}\:\frac{{x}+{y}+\frac{\pi}{\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{1}−{Ce}^{{x}} }−\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{y}=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{1}−{Ce}^{{x}} }−\mathrm{1}−\sqrt{\mathrm{2}}\right)−\frac{\pi}{\mathrm{4}}−{x} \\ $$

Commented by peter frank last updated on 22/May/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by Mr.D.N. last updated on 23/May/20

Thank you mr.w I appreciate your grand hard work but can you present your answer in form of : log{1+ tan (x+y)/2} =x+C.

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