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Question: Solution of \((x^{2} - 4xy - 2y^{2})dx + (y^{2} - 4xy - 2x^{2})dy = 0\) is...

Solution of (x2−4xy−2y2)dx+(y2−4xy−2x2)dy=0(x^{2} - 4xy - 2y^{2})dx + (y^{2} - 4xy - 2x^{2})dy = 0 is

A

x3+y3−6xy(x+y)=cx^{3} + y^{3} - 6xy(x + y) = c

B

x3+y3+6xy(x−y)=cx^{3} + y^{3} + 6xy(x - y) = c

C

x3+y3+6xy(x+y)=cx^{3} + y^{3} + 6xy(x + y) = c

D

x3+y3−6xy(x−y)=cx^{3} + y^{3} - 6xy(x - y) = c

Answer

x3+y3−6xy(x+y)=cx^{3} + y^{3} - 6xy(x + y) = c

Explanation

Solution

Comparing given equation with Mdx + Ndy = 0,

We get, M=x2−4xy−2y2M = x^{2} - 4xy - 2y^{2}, N=y2−4xy−2x2N = y^{2} - 4xy - 2x^{2}

∂M∂y=−4x−4y\frac{\partial M}{\partial y} = - 4x - 4y

∂N∂x=−4y−4x\frac{\partial N}{\partial x} = - 4y - 4x

∴ ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

So the given differential equation is exact.

Integrating m w.r.t. x, treating y as constant,

∫M⥂dx=∫(x2−4xy−2y2)dx=x33−2x2y−2y2x\int_{}^{}{M ⥂ dx} = \int_{}^{}{(x^{2} - 4xy - 2y^{2})dx} = \frac{x^{3}}{3} - 2x^{2}y - 2y^{2}xIntegrating N w.r.t. y, treating x as constant,

∫Ndy=∫(y2−4xy−2x2)dy=y33−2xy2−2x2y=y33\int_{}^{}{Ndy} = \int_{}^{}{(y^{2} - 4xy - 2x^{2})dy} = \frac{y^{3}}{3} - 2xy^{2} - 2x^{2}y = \frac{y^{3}}{3}; (omitting−2xy2−2x2y- 2xy^{2} - 2x^{2}y which already occur in ∫Mdx\int_{}^{}{Mdx})

∴ Solution of the given equation is x33−2x2y−2xy2+y33=λ\frac{x^{3}}{3} - 2x^{2}y - 2xy^{2} + \frac{y^{3}}{3} = \lambda ⇒ x3+y3−6xy(x+y)=3λx^{3} + y^{3} - 6xy(x + y) = 3\lambda

∴ x3+y3−6xy(x+y)=cx^{3} + y^{3} - 6xy(x + y) = c (3λ = c)