What is the solution for y^'+1/(20x)y=-9.81 ?

The solution for y^'+1/(20x)y=-9.81 is y=-9.34285…x+(c_{1})/(\sqrt[20]{x)}

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y^'+1/(20x)y=-9.81

y^{^{'}} + \frac{1}{20 x} y = - 9.81

y = - 9.34285 \dots x + \frac{c _{1}}{20 x}

Solution steps

y^{'} (x) + \frac{1}{20 x} y = - 9.81

Solve linear ODE:

y = - 9.34285 \dots x + \frac{c _{1}}{20 x}

y = - 9.34285 \dots x + \frac{c _{1}}{20 x}

\frac{d}{d x} (5 - cos (x))

\frac{\partial}{\partial x} (2 (x - 1)^{2} - 3 y - 5)

y^{^{'}} = x^{3} (1 - y)

y^{^{'}} = 0.001 y (1800 - y)

\int x \cdot e^{- x^{2}} d x

What is the solution for y^'+1/(20x)y=-9.81 ?
The solution for y^'+1/(20x)y=-9.81 is y=-9.34285…x+(c_{1})/(\sqrt[20]{x)}