Solve. dy/dx+3/x y = 27y^(1/3) 1n(x), x > 0

Question

Solve.

\frac{d y}{d x} + \frac{3}{x} y = 27 y^{\frac{1}{3}} 1 n (x), x > 0

Answer 1

Here the given differential equation is

y^{^{'}} + \frac{3 y}{x} = 27 y^{\frac{1}{3}}

Let

v = y^{\frac{2}{3}} t h e n \frac{d v}{d x} = \frac{2}{3} y^{\frac{1}{3}} \frac{d y}{d x}

.With respect to this transformation, the above DE reduces to

\frac{3}{2} y^{\frac{1}{3}} \frac{d v}{d x} + \frac{3 y}{x} = 27 y^{\frac{1}{3}}

\Rightarrow \frac{3}{2} \frac{d v}{d x} + \frac{3 y^{\frac{2}{3}}}{x} = 27

\Rightarrow \frac{1}{2} \frac{d v}{d x} + \frac{y^{\frac{2}{3}}}{x} = 9

\Rightarrow \frac{d v}{d x} + \frac{2 y^{\frac{2}{3}}}{x} = 18

\Rightarrow \frac{d v}{d x} + 2 \frac{v}{x} = 18

\Rightarrow x^{2} \frac{d v}{d x} + 2 x v = 18 x^{2}

\Rightarrow \frac{d}{d x} (x^{2} v) = 18 x^{2}

\Rightarrow x^{2} v = \int 18 x^{2} d x + c

\Rightarrow x^{2} v = 6 x^{3} + c

\Rightarrow v = 6 x + c x^{- 2}

\Rightarrow y^{\frac{2}{3}} = 6 x + c x^{- 2}

\Rightarrow y = {(6 x + c x^{- 2})}^{\frac{3}{2}}

Where c is arbitrary constant.

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