What is the solution for u^'=7e^{9x-\H(x)} ?

The solution for u^'=7e^{9x-\H(x)} is u(x)=integral of 7e^{9x-\H(x)}

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u^'=7e^{9x-\H(x)}

u^{^{'}} = 7 e^{9 x - H (x)}

u (x) = \int 7 e^{9 x - H (x)} d x

Solution steps

u^{'} (x) = 7 e^{9 x - H (x)}

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

then

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

u (x) = \int 7 e^{9 x - H (x)} d x

u (x) = \int 7 e^{9 x - H (x)} d x

y^{^{'}} - 2 \frac{y}{coth ( 2 x )} = \frac{- 2}{coth ( 2 x )}

e^{x} \frac{d y}{d x} = y^{2}

e^{y} d x - x^{2} d y = 0

\frac{d y}{d x} = \frac{2 x}{- 8 y - 8}, y (0) = - 8

f^{^{'}} (x) = 9 f (x)

What is the solution for u^'=7e^{9x-\H(x)} ?
The solution for u^'=7e^{9x-\H(x)} is u(x)=integral of 7e^{9x-\H(x)}