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प्रश्न
Solve the following differential equation:
`("x + a")"dy"/"dx" - 3"y" = ("x + a")^5`
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उत्तर
`(x + a)dy/dx - 3y = (x + a)^5`
∴ `dy/dx - 3y/(x + a) = (x + a)^4`
∴ `dy/dx + ((- 3)/(x + a))y = (x + a)^4` ...(1)
This is the linear differential equation of the form
`dy/dx + "P".y = "Q"`, where P = `(- 3)/(x + a)` and Q = (x + a)4
∴ I.F. = `e^(int P dx) = e^(int (- 3)/(x + a)dx) = e^(-3 int 1/(x + a)dx)`
`= e^(- 3 log |x + a|) = e^(log (x + a)^- 3)`
`= (x + a)^-3 = 1/(x + a)^3`
∴ the solution of (1) is given by
`y . ("I.F.") = int "Q" * ("I.F.") dx + c`
∴ `y * 1/(x + a)^3 = int (x + a)^4 * 1/(x + a)^3 dx + c`
∴ `y/(x + a)^3 = int (x + a) dx + c`
∴ `y/(x + a)^3 = ((x + a)^2)/2 + c`
∴ 2y = (x + a)5 + 2c (x + a)3
This is the general solution.
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