What is the solution for 3x+20yy^'=0,y(2)=5 ?

The solution for 3x+20yy^'=0,y(2)=5 is y=(sqrt(-3x^2+512))/(2sqrt(5))

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3x+20yy^'=0,y(2)=5

3 x + 20 y y^{^{'}} = 0, y (2) = 5

y = \frac{- 3 x ^{2} + 512}{2 5}

Solution steps

3 x + 20 y y^{'} (x) = 0

Solve separable ODE:

y = \frac{- 3 x ^{2} + 512}{2 5}

y = \frac{- 3 x ^{2} + 512}{2 5}

(ln (y))^{3} \frac{d y}{d x} = x^{3} y

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

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

e^{2 x} y^{^{'}} = y - y e^{3 x}

(t^{2} - 16 t + 48) \frac{d y}{d t} = y

What is the solution for 3x+20yy^'=0,y(2)=5 ?
The solution for 3x+20yy^'=0,y(2)=5 is y=(sqrt(-3x^2+512))/(2sqrt(5))