What is the solution for 16yy^'+9x=0 ?

The solution for 16yy^'+9x=0 is y=(sqrt(-9x^2+c_{1)})/4 ,y=-(sqrt(-9x^2+c_{1)})/4

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16yy^'+9x=0

16 y y^{^{'}} + 9 x = 0

y = \frac{- 9 x ^{2} + c _{1}}{4}, y = - \frac{- 9 x ^{2} + c _{1}}{4}

Solution steps

16 y y^{'} (x) + 9 x = 0

Solve separable ODE:

y = \frac{- 9 x ^{2} + c _{1}}{4}, y = - \frac{- 9 x ^{2} + c _{1}}{4}

y = \frac{- 9 x ^{2} + c _{1}}{4}, y = - \frac{- 9 x ^{2} + c _{1}}{4}

x y^{^{'}} = y + 1 + x y + x

y^{^{'}} = \frac{sin ( 2 x )}{cos ( 3 y )}, y (0) = 1

y^{^{'}} (t) = (3 + y) (1 - y), y (0) = 4

2 y t (d y) + y^{2} (d t) = 0

y^{^{'}} + y = y (t e^{t^{2}} + 1), y (0) = 1

What is the solution for 16yy^'+9x=0 ?
The solution for 16yy^'+9x=0 is y=(sqrt(-9x^2+c_{1)})/4 ,y=-(sqrt(-9x^2+c_{1)})/4