Question

If y² = a²cos²x + b²sin²x, show that `y + (d^2y)/(dx^2) = (a^2b^2)/y^3`

Answer 1

y² = a²cos²x + b²sin²x                           ...(1)
Differentiating both sides w.r.t. x, we get

`2y"dy"/"dx" = a^2"d"/"dx"(cosx)^2 + b^2"d"/"dx"(sinx)^2`

= `a^2 xx 2cosx."d"/"dx"(cosx) + b^2 xx 2sinx."d"/'dx"(sinx)`

= a² x 2 cos x (– sin x) + b² x 2 sin x cos x
= (b² – a²) sin2x

∴ `y"dy"/"dx" = ((b^2 - a^2)/2)sin2x`    ...(2)

Differentiating again w.r.t. x, we get

`y."d"/"dx"(dy/dx) + "dy"/"dx"."dy"/"dx" = ((b^2 - a^2)/2)."d"/"dx"(sin2x)`

∴ `y(d^2y)/(dx^2) + (dy/dx)^2 = ((b^2 - a^2)/2) xx cos2x xx 2`

∴ `y(d^2y)/(dx^2) + (dy/dx)^2 = (b^2 - a^2)cos2x`

∴ `y^3(d^2y)/(dx^2) +y^2 (dy/dx)^2 = y^2(b^2 - a^2)cos2x`

∴ `y^3(d^2y)/(dx^2) = y^2(b^2 - a^2)cos2x - y^2(dy/dx)^2`

∴ `y^4 + y^3(d^2y)/(dx^2) = y^2(b^2 - a^2)cos2x - y^2(dy/dx)^2 + y^4`

= (a²cos²x + b²sin²x)(b² – a²)(cos²x – sin²x) – [(b² – a²)sin x cosx]² + (a²cos²x + b²sin²x)²    ...[By (1) and (2)]

= (a²b²cos²x – a⁴cos²x + b⁴sin²x – a²b²sin²x) x (cos²x –sin²x) – (b⁴sin²xcos²x + a⁴sin²xcos²x  – 2a²b²sin²x cos²x) + (a⁴cos⁴x + b⁴sin⁴x + b⁴sin⁴x + 2a²b²sin²x cos²x)

= a²b²cos4x – a²b²sin²xcos²x – a⁴cos⁴x + a⁴sin²xcos²x + b⁴sin²xcos²x – b⁴sin²xcos²x – a⁴sin²xcos²x + 2a²b²sin²xcos²x + a⁴cos⁴x + b⁴x + b⁴sin⁴x + 2a²b²sin²xcos²x

= a²b²cos⁴x + 2a²b²sin²xcos²x + a²b²sin⁴x
= a²b² (sin⁴x + 2sin²x cos²x + cos⁴x)

∴ `y^4 + y^3(d^2y)/(dx^2)` = a²b²      ...[∵ sin²x + cos²x = 1]

∴ `y^3(y + (d^2y)/(dx^2))` = a²b² 

∴ `y + (d^2y)/(dx^2) = (a^2b^2)/(y^3)`.