What is the solution for u^'=e^{4x-\H(x)} ?

The solution for u^'=e^{4x-\H(x)} is u(x)=integral of e^{4x-\H(x)}

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u^'=e^{4x-\H(x)}

u^{^{'}} = e^{4 x - H (x)}

u (x) = \int e^{4 x - H (x)} d x

Solution steps

u^{'} (x) = e^{4 x - H (x)}

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

then

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

u (x) = \int e^{4 x - H (x)} d x

u (x) = \int e^{4 x - H (x)} d x

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

\frac{d y}{d x} + 5 y = 20, y (0) = 2

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

y^{^{'}} = x^{2} - 2 x^{2} y

1 - y^{2} (d x) - 1 + x^{2} (d y) = 0

What is the solution for u^'=e^{4x-\H(x)} ?
The solution for u^'=e^{4x-\H(x)} is u(x)=integral of e^{4x-\H(x)}