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

Answer:

$tan^{-1}(y) = -tan^{-1} e^{x} +  {\pi/2}$,

Step-by-step explanation:

Hi,

Given equation is

$(1 + e^{2x})dy + ( 1 + y^{2})e^{x}dx = 0$

Separating x terms to one side and y terms to other,

we get

$\dfrac{dy}{( 1 + y^{2})} = -\dfrac{e^{x} dx}{(1 + e^{2x})}$,

Integrating on both sides, we get

$\int \dfrac{dy}{( 1 + y^{2})} = -\int \dfrac{e^{x} dx}{(1 + e^{2x})}$,

${tan^{-1} (y)} = -{tan^{-1} e^{x}} + c$,

Given that y = 1, when x = 0,

hence, we get

${tan ^{-1}(1)} = -{tan^{-1} e^{0}} + c$,

π/4 = -π/4 + c

c = π/2,

Hence, the particular solution is

${tan^{-1} (y)} = -{tan ^{-1}e^{x}} + {\pi/2}$,

Hope, it helps !