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Question
Solve the following Linear differential equation:
`("d"y)/("d"x) + (3y)/x = 1/x^2`, given that y = 2 when x = 1
Solution
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.
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