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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Solve the differential equation $\large\frac{dy}{dx}$$=cos(x+y)+sin(x+y)$ and find the particular solution if $y(0)=0$

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Toolbox:
  • $sinx=\large\frac{2tan\large\frac{x}{2}}{1+tan^2\frac{x}{2}}$
  • $cosx=\large\frac{1-tan^2\large\frac{x}{2}}{1+tan^2\large\frac{x}{2}}$
  • $1+tan^2\large\frac{x}{2}=sec^2\large\frac{x}{2}$
$\large\frac{dy}{dx}$$=sin(x+y)+cos(x+y)$
Let $x+y=z$
$\Rightarrow\:\large\frac{dy}{dx}=\frac{dz}{dx}$$-1$
Substituting the the values in the equation
$\large\frac{dz}{dx}$$-1=sinz+cosz$
$\Rightarrow\:\large\frac{dz}{dx}$$=sinz+cosz+1$
$\Rightarrow\:\large\frac{dz}{1+sinz+cosz}$$=dx$
$\Rightarrow\:\int\large\frac{dz}{1+sinz+cosz}$$=\int dx$
$1+sinz+cosz=1+\large\frac{2tan\large\frac{z}{2}}{1+tan^2\frac{z}{2}}+\large\frac{1-tan^2\large\frac{z}{2}}{1+tan^2\large\frac{z}{2}}$
$\Rightarrow\:1+sinz+cosz=\large\frac{1+tan^2\frac{z}{2}+2tan\frac{z}{2}+1-tan^2\frac{z}{2}}{1+tan^2\large\frac{z}{2}}$
$\Rightarrow\:1+sinz+cosz=\large\frac{2+2tan\frac{z}{2}}{1+tan^2\large\frac{z}{2}}=\frac{2(1+tan\large\frac{z}{2})}{sec^2\frac{z}{2}}$
$\Rightarrow\:\int\large\frac{dz}{1+sinz+cosz}$$=\int dx$
$\Rightarrow\:\int \large\frac{sec^2\large\frac{z}{2}}{2(1+tan\frac{z}{2})}$$dz=\int dx$
Put $tan\large\frac{z}{2}$$=t$
Differentiating both the sides,
$\large\frac{1}{2}$$sec^2 \large\frac{z}{2}$$dz=dt$
$\Rightarrow\:\int \large\frac{dt}{1+t}$$=x+c$
$\Rightarrow\:log|1+t|=x+c$
substituting the value of $t$
$log|1+tan\large\frac{z}{2}$$|=x+c$
Again substituting the value of $z$,
$log|1+tan\large\frac{x+y}{2}$$|=x+c$
Given: $ y(0)=0$ $\Rightarrow\:x=y=0$
Substituting the values of $x\:\;and \:\:y$
$c=0$
$\therefore\:$ The solution is $log|1+tan\large\frac{x+y}{2}$$|=x$
answered Feb 6, 2014 by rvidyagovindarajan_1
 

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