Q9.    Find the particular solution of the differential equation (1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0, given that y = 1 when x = 0.

Answers (1)
G Gautam harsolia

Given equation is
(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0
we can rewrite it as
\frac{dy}{dx}= -\frac{(1+y^2)e^x}{(1+e^{2x})}\\ \\ \frac{dy}{1+y^2}= \frac{-e^xdx}{1+e^{2x}}
Now, integrate both the sides
\tan^{-1}y + C' =\int \frac{-e^{x}dx}{1+e^{2x}}
\int \frac{-e^{x}dx}{1+e^{2x}}\\
Put
 e^x = t \\ e^xdx = dt
\int \frac{dt}{1+t^2}= \tan^{-1}t + C''
Put t = e^x again
\int \frac{-e^{x}dx}{1+e^{2x}} = -\tan ^{-1}e^x+C''
Put this in our equation
\tan^{-1}y = -\tan ^{-1}e^x+C\\ \tan^{-1}y +\tan ^{-1}e^x=C
Now, by using boundary conditions we will find the value of C
It is given that
y = 1 when x = 0
\\ \tan^{-1}1 +\tan ^{-1}e^0=C\\ \\ \frac{\pi}{4}+\frac{\pi}{4}= C\\ C = \frac{\pi}{2}
Now, put the value of C

\tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}
Therefore,  the particular solution of the differential equation (1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0   is   \tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}

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