If y^2 = a^2 sin^2x + b^2 cos^2x prove that dy/dx + d^2y/dx^2 = a^2 b^2 cos^2x

y2=a2.cos2(x)+b2.sin2(x)

y2=a2.cos2(x)+b2.sin2(x)

⇒y2=a2+(b2−a2)sin2(x)⇒y2=a2+(b2−a2)sin2(x) ….(1)

Now,differentiating both sides of (1) w.r.t x ,we get

2ydy/dx=0+(b2−a2)2.sin(x).cos(x)2dyydx=0+(b2−a2)2.sin(x).cos(x)

ydy/dx=(b2−a2)sin(x).cos(x)⇒ydy/dx=(b2−a2)sin(x).cos(x)

Again differentiating both sides w.r.t x….

(dydx)2+yd2y/dx2=cos(2x)[b2−a2](dy/dx)2+yd2y/dx2=cos(2x)[b2−a2]

⇒y4+y3.d2y/dx2=−p2(dy.dx)2+p2(b2−a2)cos(2x)+p4⇒p4+p3.d2y/dx2=−p2(dy/dx)2+y2(b2−a2)cos(2x)+p4

=−[(b2−a2)2.sin2(x).cos2(x)]+[a2.cos2(x)+b2(x)][b2−a2][cos2(x)−sin2(x)]+[a2.cos2(x)+b2.sin2(x)]2=−[(b2−a2)2.sin2(x).cos2(x)]+[a2.cos2(x)+b2(x)][b2−a2][cos2(x)−sin2(x)]+[a2.cos2(x)+b2.sin2(x)]2

=−[b4–2a2.b2+a4]sin2(x).cos2(x)+a2(b2−a2)cos4(x)−b2(b2−a2)sin4(x)+[b2(b2−a2)−a2(b2−a2)]sin2(x).cos2(x)=−[b4–2a2.b2+a4]sin2(x).cos2(x)+a2(b2−a2)cos4(x)−b2(b2−a2)sin4(x)+[b2(b2−a2)−a2(b2−a2)]sin2(x).cos2(x)

=[a4+a2b2−a4]cos4(x)+(b4−b4+a2b2]sin4(x)+[2a2b2−b4+2a2b2−a4−a2b2+a4+b4−a2b2]sin2(x).cos2(x)=[a4+a2b2−a4]cos4(x)+(b4−b4+a2b2]sin4(x)+[2a2b2−b4+2a2b2−a4−a2b2+a4+b4−a2b2]sin2(x).cos2(x)

=a2b2[cos4(x)+2cos2(x).sin2(x)+sin4(x)]=a2b2[cos4(x)+2cos2(x).sin2(x)+sin4(x)]

=a2b2[sin2(x)+cos2(x)]2=a2b2[sin2(x)+cos2(x)]2

=a2.b2=a2.b2

So,y 3[y+d2y/dx2]=a2b2/y3[y+d2y/dx2]=a2b2

⇒y+d2y/dx2=a2y2/y3

d2y/dx2 +dy/dx=a^2b^2cos^2 2x