Question

If log y = log (sin x) – x², show that `(d^2y)/(dx^2) + 4x "dy"/"dx" + (4x^2 + 3)y` = 0.

Answer 1

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.