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प्रश्न
Show that y = `"a"x + "b"/x ≠ 0` is a solution of the differential equation x2yn + xy’ – y = 0
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उत्तर
Given y = `"a"x + "b"/x`
y = `("a"x^2 + "b")/x`
xy = ax2 + b .......(1)
DIfferentiating equation (1) w.r.t 'x', we get
`x ("d"y)/("d"x) + y(1)` = a 2x + 0
`x ("d"y)/("d"x) + y` = 2ax
2a = `(x ("d"y)/("d"x) + y)/x `
= `(xy"'" + y)/x`
Again differentiating we get
`x ("d"^2y)/("d"x^2) + ("d"y)/("d"x) (1) + ("d"y)/("d"x)` = 2a
`x ("d"^2y)/("d"x^2) + ("d"y)/("d"x) + ("d"y)/("d"x) = (x ("d"y)/("d"x) + y)/x`
`x ("d"^2y)/("d"x^2) + 2 ("d"y)/("d"x) - (x ("d"y)/("d"x) + y)/x` = 0
Multiply by 'x'
`x^2 ("d"^2y)/("d"x^2) + 2x ("d"y)/("d"x) - (x ("d"y)/("d"x) + y)` = 0
`x^2 ("d"^2y)/("d"x^2) + 2x ("d"y)/("d"x) - x ("d"y)/("d"x) - y` = 0
x2y” + 2xy’ – xy’ – y = 0
x2y” + xy’ – y = 0
∴ y = `"a"x + "b"/x` is a solution of the differential equation x2y” + xy’ – y = 0
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