What is the solution for y^'=-(9x)/(16y) ?

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

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y^'=-(9x)/(16y)

y^{^{'}} = - \frac{9 x}{16 y}

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

Solution steps

y^{'} (x) = - \frac{9 x}{16 y}

Solve separable ODE:

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

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

y^{^{''}} + y = 3 e^{- 8 x}, y (0) = 0, y^{^{'}} (0) = 0

x \to 4 lim (16 - x^{2})

x \to - 2 lim (5)

y = x^{3} - 4 x + 3

\int 4 x + 5 d x

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