If u= tan^-1(y²/x) then prove that x² d²u/dx²+2xy d²u/dxdy+y² d²u/dy²=-sin²u•sin2u
Answers
Answer:
Correct option is A)
Given : u=tan
−1
(
x−y
x
3
+y
3
)
tanu=
x−y
x
3
+y
3
(x−y)tanu=x
3
+y
3
→(1)
Differentiate w.r.t
′
x
′
partially
+tanu+(x−y)sec
2
u
∂x
∂u
=3x
2
→(2)
Differntiate w.r.t
′
y
′
partially
−tanu−(x−y)sec
2
u
∂x
∂u
=3x
2
→(3)
Further x(2) + y(3)
(x−y)tan(u)+sec
2
u(x−y){x
∂x
∂u
+y
∂y
∂u
}=3(x
3
+y
3
)
(x−y)[tan(u)+sec
2
u{x
∂x
∂u
+y
∂y
∂u
}]=3(x
3
+y
3
)
tan(u)+sec
2
u{x
∂x
∂u
+y
∂y
∂u
}=
(x−y)
3(x
3
+y
3
)
tan(u)+sec
2
u{x
∂x
∂u
+y
∂y
∂u
}=3tanu
⇒x
∂x
∂u
+y
∂y
∂u
=sin(2u)
Differentiating w.r.t x and y gives
∂x
∂u
+x
∂x
2
∂
2
u
+x
∂x∂y
∂
2
u
=2cos(2u)
∂x
∂u
→(4)
x
∂x∂y
∂
2
u
+y
∂y
2
∂
2
u
+
∂y
∂u
=2cos(2u)
∂y
∂u
Further x(4) + y(5) gives
x
2
∂x
2
∂
2
u
+2xy
∂x∂y
∂
2
u
+y
2
∂y
2
∂
2
u
={2cos(2u)−1}{x
∂x
∂u
+y
∂y
∂u
}
={2cos(2u)−1}sin(2u)