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# Find the particular solution of the differential equation $(1+e^{2x})dy + (1+y^2)e^xdx$ = 0, given that $y = 1$ when $x = 0$

Toolbox:
• The linear differential equation of the form $\large\frac{dy}{dx} $$= F(x,y), where F(x,y) is of the form g(x).h(y), where g(x) is a function with x and h(y) is a function with y, then this type of equation is called variables seperable and the the equation can be solved seperating the variables. Step 1: Given : (1+e^{2x})dy+(1+y^2)e^xdx=0 The equation can be written as \large\frac{dy}{dx} = \large\frac{-e^x(1+y^2)}{(1+e^2x)} Using the information in the tool box, we understand that this equation is variables seperable and we can solve this by seperating the variables. Seperating the variables we get, \large\frac{dy}{(1+y^2)} = \large\frac{-e^xdx}{(1+e^2x)} Step 2: Integrating on both sides we get, \int\large\frac{dy}{(1+y^2)}=\int\large\frac{-e^xdx}{(1+e^{2x})} \tan^{-1}y=-\int\large\frac{ e^xdx}{(1+e^{2x})} \int\large\frac{ e^xdx}{(1+e^{2x})} can be solved by method of substitution, Let e^x = t; then dt = e^xdx Substituting this we get, \int\large\frac{dt}{(1+t^2)}$$ = \tan^{-1}t$
Substituting for t we get,
$\tan^-1(e^x)$
$\tan^{-1}y=-\tan^{-1}(e^x)+C$
$\tan^{-1}x+\tan^{-1}y=C$
Step 3:
To evaluate the value $C$,let us substitute the given values of $x=0$ and $y=1$
$\tan^{-1}x+\tan^{-1}e^0=C$
$\tan^{-1}1=\large\frac{\pi}{4}$
$C=\large\frac{\pi}{4}+\frac{\pi}{4}$
$C=\large\frac{\pi}{2}$
Substituting this we get,
$\tan^{-1}x + \tan^{-1}y = \large\frac{\pi}{2}$
edited Jul 30, 2013