Find a particular solution of the differential equation (x + 1) dy/dx = 2 e⁻ʸ − 1, given that y = 0 when x = 0

 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 

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