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Solve: (x^3 + 3xy^2) dx + (y^3 + 3x^2y)dy = 0

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Solve: $(x^{3} + 3 x y^{2}) d x + (y^{3} + 3 x^{2} y) d y = 0$

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(x^{3} + 3 x y^{2}) d x + (y^{3} + 3 x^{2} y) d y = 0

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Text solutionVerified

\Rightarrow \frac{d y}{d x} = - \frac{( x ^{3} + 3 x y ^{2} )}{( y ^{3} + 3 x ^{2} y )} = - \frac{3 x y ^{2} ( \frac{x ^{3}}{3 x y ^{2}} + 1 )}{3 x ^{2} y ( \frac{y ^{3}}{3 x ^{2} y} + 1 )} = - \frac{y ( \frac{x ^{2}}{3 y ^{2}} + 1 )}{x ( \frac{y ^{2}}{3 x ^{2}} + 1 )}

\Rightarrow \frac{d y}{d x} = f (\frac{y}{x})

\Rightarrow

the given differential equation is a homogeneous equation.

The solution of the given differential equation is :

Put

y = vx

\Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}

v + x \frac{d v}{d x} = - \frac{vx ( \frac{x ^{2}}{3 ( vx ) ^{2}} + 1 )}{x ( \frac{( vx ) ^{2}}{3 x ^{2}} ) + 1} = - v \frac{( \frac{1}{3 ( v ) ^{2}} + 1 )}{( \frac{( v ) ^{2}}{3} + 1 )} = - \frac{1 + 3 ( v ) ^{2}}{3 + ( v ) ^{2}} \times \frac{1}{v}

= - \frac{1 + 3 ( v ) ^{2}}{3 v + ( v ) ^{3}}

\Rightarrow x \frac{d v}{d x} = - \frac{1 + 3 ( v ) ^{2}}{3 v + ( v ) ^{3}} - v = - \frac{1 + 3 ( v ) ^{2} + 3 ( v ) ^{2} + ( v ) ^{4}}{3 v + ( v ) ^{3}} = \frac{1 + 6 ( v ) ^{2} + ( v ) ^{4}}{3 v + ( v ) ^{3}}

\Rightarrow \frac{3 v + ( v ) ^{3}}{1 + 6 ( v ) ^{2} + ( v ) ^{4}} d v = - \frac{d x}{x}

Integrating both the sides we get:

\Rightarrow \int \frac{3 v + ( v ) ^{3}}{1 + 6 ( v ) ^{2} + ( v ) ^{4}} d v = - \int \frac{d x}{x} + c

As we know that,

\frac{d}{d v} (1 + 6 (v)^{2} + (v)^{4}) = 12 v + 4 v^{3} = 4 (3 v + v^{3})

\Rightarrow \frac{l n ∣1 + 6 ( v ) ^{2} + ( v ) ^{4}}{4} + l n ∣ x ∣ = l n ∣ c ∣

Resubstituting the value of

y = vx

we get

\Rightarrow \frac{l n ∣1 + 6 ( \frac{y}{x} ) ^{2} + ( \frac{y}{x} ) ^{4}}{4} + l n ∣ x ∣ = l n ∣ c ∣

\Rightarrow y^{4} + 6 x^{2} y^{2} + x^{3} 4 = C

Ans :

\Rightarrow y^{4} + 6 x^{2} y^{2} + x^{3} 4 = C

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