Prove that $x^{2}-y^{2}=C\left(x^{2}+y^{2}\right)^{2}$ is the general solution of the differential equation $\left(x^{3}-3xy^{2}\right)dx=\left(y^{3}-3x^{2}y\right)dy$, where $C$ is a parameter.

CBSE Class 12 · 2017 · Differential EquationsLong Answer

Prove that

x^{2} - y^{2} = C (x^{2} + y^{2})^{2}

is the general solution of the differential equation

(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y

, where

C

is a parameter.

Answer: Proved

It is enough to show the given relation satisfies the differential equation. From

x^{2} - y^{2} = C (x^{2} + y^{2})^{2}

, write

C = \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}}

and differentiate the relation:

2 x - 2 y y^{'} = C \cdot 2 (x^{2} + y^{2}) (2 x + 2 y y^{'})

Substitute

C

and cancel

(x^{2} + y^{2})

(x - y y^{'}) (x^{2} + y^{2}) = 2 (x^{2} - y^{2}) (x + y y^{'})

Expanding both sides and collecting terms:

- x^{3} + 3 x y^{2} - 3 x^{2} y y^{'} + y^{3} y^{'} = 0

\Rightarrow (x^{3} - 3 x y^{2}) = (y^{3} - 3 x^{2} y) y^{'}

which is exactly

(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y

. Hence the relation is the general solution.

Proved

CBSE Class 12 Board · 2017 · Subjective