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Question
Solve the following differential equation:
x dx + 2y dx = 0, when x = 2, y = 1
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Solution
x dx + 2y dx = 0
∴ x dy = - 2y dx
∴ `1/"y" "dy" = (- 2)/"x" "dx"`
Integrating, we get
`int 1/"y" "dy" = - 2 int 1/"x" "dx"`
∴ log |y| = - 2 log |x| + log c
∴ log |y| = - log |x2| + log c
∴ log |y| = log `|"c"/"x"^2|`
∴ y = `"c"/"x"^2`
∴ x2y = c
This is the general solution.
When x = 2, y = 1, we get
4(1) = c
∴ c = 4
∴ the particular solution is
x2y = 4.
Notes
The answer in the textbook is incorrect.
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