What is the solution for y^'=(2y)/(3x),y(27)=4 ?

The solution for y^'=(2y)/(3x),y(27)=4 is y=(4x^{2/3})/9

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y^'=(2y)/(3x),y(27)=4

y^{^{'}} = \frac{2 y}{3 x}, y (27) = 4

y = \frac{4 x ^{\frac{2}{3}}}{9}

Solution steps

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

Solve separable ODE:

y = \frac{4 x ^{\frac{2}{3}}}{9}

y = \frac{4 x ^{\frac{2}{3}}}{9}

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What is the solution for y^'=(2y)/(3x),y(27)=4 ?
The solution for y^'=(2y)/(3x),y(27)=4 is y=(4x^{2/3})/9