Question

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

Answer 1

Given, differential equation is $(1+e^{2x})dy+(1+y^2)e^{x}dx=0$
Separating the variables, we get $\frac{dy}{1+y^2}+\frac{e^{x}dx}{1+e^{2x}}=0$
On integrating both sides, we get $\int \frac{dy}{1+y^2}+\int \frac{e^{x}dx}{1+e^{2x}}=C$
Put $t=e^x\Rightarrow e^{x}dx=dt\Rightarrow tany+\int \frac{dt}{1+t^2}=C$
$\Rightarrow ta{n}^{-1}y+ta{n}^{-1}t=C\Rightarrow ta{n}^{-1}y+ta{n}^{-1}{e}^x=C\dots \dots \left(i\right)$
Now, put x = 0 and y = 1
$\therefore ta{n}^{-1}+ta{n}^{-1}{e}^0=C\Rightarrow \frac{\pi }{4}+\frac{\pi }{4}=C\Rightarrow C=\frac{\pi }{2}$
On putting the value of C in Eq. (i), we get $ta{n}^{-1}y+ta{n}^{-1}{e}^x=\frac{\pi }{2}$
which is the required particular solution.