What is the solution for u^'-2/x \H(x)=(-2)/x ?

The solution for u^'-2/x \H(x)=(-2)/x is u(x)=integral of (-2+2\H(x))/x

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u^'-2/x \H(x)=(-2)/x

u^{^{'}} - \frac{2}{x} H (x) = \frac{- 2}{x}

u (x) = \int \frac{- 2 + 2 H ( x )}{x} d x

Solution steps

u^{'} (x) - \frac{2}{x} H (x) = \frac{- 2}{x}

Isolate

u^{'} (x) : u^{'} (x) = \frac{- 2 + 2 H ( x )}{x}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

u (x) = \int \frac{- 2 + 2 H ( x )}{x} d x

u (x) = \int \frac{- 2 + 2 H ( x )}{x} d x

(x^{2} + 16) \frac{d y}{d x} + x y = 2 x

\frac{d y}{d x} = 6 x 5 y + 7, y (0) = 9

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}, y (1) = ln (11)

5 y^{^{'}} - 6 e^{3 x} y = 0, y (0) = 5

y^{^{'}} + \frac{y}{10} = \frac{v}{10}

What is the solution for u^'-2/x \H(x)=(-2)/x ?
The solution for u^'-2/x \H(x)=(-2)/x is u(x)=integral of (-2+2\H(x))/x