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Question
If log y = log (sin x) – x2, show that `(d^2y)/(dx^2) + 4x "dy"/"dx" + (4x^2 + 3)y` = 0.
Solution
log y = log (sin x) – x2
∴ log y = `log (sin x) – log e^(x^2)`
∴ log y = `log(sinx/e^(x^2))`
∴ y = `sinx/e^(x^2)`
∴ `e^(x^2).y` = sin x ...(1)
Differentiating both sides w.r.t. x, we get
`e^(x^2)."dy"/"dx" + y."d"/"dx"e^(x^2) = "d"/"dx"(sinx)`
∴ `e^(x^2)."dy"/"dx" + y.e^(x^2)."d"/"dx"(x^2)` = cos x
∴ `e^(x^2)."dy"/"dx" + y.e^(x^2) xx 2x` = cos x
∴ `e^(x^2)("dy"/"dx" + 2xy)` = cos x
Differentiating again w.r.t. x, we get
`e^(x^2)."d"/"dx"("dy"/"dx" + 2xy) + ("dy"/"dx" + 2xy)."d"/"dx"(e^(x^2)) = "d"/"dx"(cosx)`
∴ `e^(x^2)[(d^2y)/(dx^2) + 2(x"dy"/"dx" + y xx 1)] + ("dy"/"dx" + 2xy).e^(x^2)."d"/"dx"(x^2)` = – sin x
∴ `e^(x^2)[(d^2y)/(dx^2) + 2x"dy"/"dx" + y] + ("dy"/"dx" + 2xy).e^(x^2) xx 2x` = – sin x
∴ `e^(x^2)[(d^2y)/(dx^2) + 2x"dy"/"dx" + 2y + 2x"dy"/"dx" + 4x^2y]`
= `-e^(x^2).y` ...[By (1)]
∴ `(d^2y)/(dx^2) + 4x"dy"/"dx" + 4x^2y + 2y` = – y
∴ `(d^2y)/(dx^2) + 4x"dy"/"dx" + 4x^2y + 2y + y` = 0
∴ `(d^2y)/(dx^2) + 4x"dy"/"dx" + 4x^2y + 3y` = 0
∴ `(d^2y)/(dx^2) + 4x"dy"/"dx" + (4x^2 + 3)y` = 0.
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