What is the solution for u^'=2e^{3x-\H(x)} ?

The solution for u^'=2e^{3x-\H(x)} is u(x)=integral of 2e^{3x-\H(x)}

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u^'=2e^{3x-\H(x)}

u^{^{'}} = 2 e^{3 x - H (x)}

u (x) = \int 2 e^{3 x - H (x)} d x

Solution steps

u^{'} (x) = 2 e^{3 x - H (x)}

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

then

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

u (x) = \int 2 e^{3 x - H (x)} d x

u (x) = \int 2 e^{3 x - H (x)} d x

\frac{d y}{d x} - 8 \frac{d y}{d x} + 16 y = 0

y^{^{'}} = x (6 - y)

\frac{d y}{d x} = \frac{3 x + 3 x y ^{2}}{y x ^{2} + 2 y}

x y^{^{'}} + 4 y = 0, y (1) = \frac{3}{2}

\frac{d y}{d x} = x (y - 1) - 2 y + 2

What is the solution for u^'=2e^{3x-\H(x)} ?
The solution for u^'=2e^{3x-\H(x)} is u(x)=integral of 2e^{3x-\H(x)}