What is the solution for u^'=3e^{8x-\H(x)} ?

The solution for u^'=3e^{8x-\H(x)} is u(x)=integral of 3e^{8x-\H(x)}

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u^'=3e^{8x-\H(x)}

u^{^{'}} = 3 e^{8 x - H (x)}

u (x) = \int 3 e^{8 x - H (x)} d x

Solution steps

u^{'} (x) = 3 e^{8 x - H (x)}

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

then

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

u (x) = \int 3 e^{8 x - H (x)} d x

u (x) = \int 3 e^{8 x - H (x)} d x

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

(\frac{y}{sec ^{2} ( x )}) \frac{d y}{d x} = 2 tan (x) + 1

x y d y = (1 + y^{2}) d x, y (2) = 3

0.5 y^{^{'}} + 50 y = 30

y - 3 y^{^{'}} = 1

What is the solution for u^'=3e^{8x-\H(x)} ?
The solution for u^'=3e^{8x-\H(x)} is u(x)=integral of 3e^{8x-\H(x)}