Find a particular solution of the differential equation
(x + 1) dy/dx = 2 e⁻ʸ − 1,
given that y = 0 when x = 0
variable separable differential equation with initial condition
given
(x + 1) dy/dx = 2 e⁻ʸ − 1,
separating
dy / (2 e⁻ʸ − 1) = dx / (x + 1)
multiply numerator and denominator of LHS with eʸ
eʸ dy / (2 − eʸ) = dx / (x + 1)
integrating
∫ eʸ / (2 − eʸ) dy = ∫ 1 / (x + 1) dx
for the LHS use the substitution
u = 2 − eʸ
⇒ du = −eʸ dy
eʸ dy =-du
we get
∫ −du / u = ∫ 1 / (x + 1) dx
−ln|u| = ln|x + 1| + C
re substitute the value of u = 2 − eʸ
−ln|2 − eʸ| = ln|x + 1| + C
ln|2 − eʸ| +ln|x + 1| = -C
using properties of logarithms,
ln|(2 − eʸ)(x + 1)| = -C
using definition of logarithm
(x + 1)(2 − eʸ) = A-----(1)
, where A is another arbitrary constant
use the initial condition y = 0 when x = 0 to solve for A
we get A =1
(1)⇒ (x + 1)(2 − eʸ) =1
or
2 − eʸ = 1 / (x + 1)
eʸ = 2 − [ 1 / (x + 1)]
eʸ = (2x + 1) / (x + 1)
using definition of log
y = ln[(2x + 1) / (x + 1)]
for more details watch this video
cbse 12th past year question papers 2025 2026 differential equations
