Advertisements
Advertisements
प्रश्न
Solve the following Linear differential equation:
`("d"y)/("d"x) + (3y)/x = 1/x^2`, given that y = 2 when x = 1
उत्तर
The given differential equation can be written as
`("d"y)/("d"x) + (3y)/x = 1/x^2`
This is of the form `("d"y)/("d"x) + "P"y` = Q
Where P = `3/x`
Q = `1/x^2`
Thus, the given differential equation is linear.
I.F = `"e"^(int "Pd"x)`
= `"e"^(int 3/4 "d"x)`
= `"e"^(3logx)`
= `"e"^(logx3)`
= x3
So, its solution is given by
y × I.F = `("Q" xx "I.F") "d"x + "c"`
yx3 = `int 1/x^2 + x^3 "d"x + "c"`
= `int x "d"x + "c"`
yx3 = `x^2/2 + "c"`
2yx3 = x2 + c
Given that y = 2 when x = 1
2 (2) (1)3 = 1 + c
4 – 1 = c
c = 3
∴ 2yx3 = x2 + 3 is a required solution.
APPEARS IN
संबंधित प्रश्न
Solve the following Linear differential equation:
`cos x ("d"y)/("d"x) + y sin x ` = 1
Solve the following Linear differential equation:
`(1 - x^2) ("d"y)/("d"x) - xy` = 1
Solve the following Linear differential equation:
`("d"y)/("d"x) + y/x = sin x`
Solve the following Linear differential equation:
`(x^2 + 1) ("d"y)/("d"x) + 2xy = sqrt(x^2 + 4)`
Solve the following Linear differential equation:
(2x – 10y3)dy + y dx = 0
Solve the following Linear differential equation:
`x sin x ("d"y)/("d"x) + (x cos x + sin x)y = sinx`
Solve the following Linear differential equation:
`(y - "e"^(sin^-1)x) ("d"x)/("d"y) + sqrt(1 - x^2)` = 0
Solve the following Linear differential equation:
`("d"y)/("d"x) + y/((1 - x)sqrt(x)) = 1 - sqrt(x)`
Solve the following Linear differential equation:
`(1 + x + xy^2) ("d"y)/("d"x) + (y + y^3)` = 0
Solve the following Linear differential equation:
`("d"y)/("d"x) + y/(xlogx) = (sin2x)/logx`
Solve the following Linear differential equation:
`(x + "a") ("d"y)/("d"x) - 2y = (x + "a")^4`
Solve the following Linear differential equation:
`x ("d"y)/("d"x) + y = x log x`
Solve the following Linear differential equation:
`x ("d"y)/("d"x) + 2y - x^2 log x` = 0
Choose the correct alternative:
Integrating factor of the differential equation `("d"y)/("d"x) = (x + y + 1)/(x + 1)` is
