ncert cbse mathematics class 12th differential equations
exercise 9.4 variable separable differential equations type
3. solve (dy/dx)+y=1
(dy/dx)+y=1
(dy/dx)=1-y
separating the variables
dy / [1-y] =dx
integrating,
[remember to divide by (-1), the coefficient of y]
{ln[1-y] }/(-1)=x+ln C
{ here ln refers to natural logarithms }
-ln(1-y) =x+ln C
lnC +ln(1-y) = -x
using properties of logarithms
ln [C(1-y)]= (-x)
getting rid of logarathmic funtion
C(1-y)= e^(-x)
(1-y) = (1/C) e^(-x)
y=1-(1/C) e^(-x)
put (1/C) =(-A)
y = 1 + A e^(-x)
4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0
divide each term by tan y tan x
[ { (sec x)^2}/ tanx ]dx + [ { (sec y)^2}/ tan y ]dy = 0
integrating term by term
keeping in mind that integral of {(f ') / f }form is ln(f)
{ here ln refers to natural logarithms }
ln(tan x) +ln(tan y) =lnC
using properties of logarithms
ln(tanx tany ) =lnC
tan x tan y = C
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ncert cbse 12th mathematics
exercise 9.4
differential equations
variable separable type of differential equations
solve dy/dx = [1-cosx]/[1+cosx]
2. solve (dy/dx) = sqrt[4-(y^2)]
3. solve (dy/dx)+y=1
4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0
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