What is the solution for y^'x+4yx=8,y(0)=1 ?

The solution for y^'x+4yx=8,y(0)=1 is y=(8\Ei(4x))/(e^{4x)}+(1-8\Ei(4*0))/(e^{4x)}

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y^'x+4yx=8,y(0)=1

y^{^{'}} x + 4 y x = 8, y (0) = 1

y = \frac{8 Ei ( 4 x )}{e ^{4 x}} + \frac{1 - 8 Ei ( 4 \cdot 0 )}{e ^{4 x}}

Solution steps

y^{'} (x) x + 4 y x = 8

Solve linear ODE:

y = \frac{8 Ei ( 4 x )}{e ^{4 x}} + \frac{1 - 8 Ei ( 4 \cdot 0 )}{e ^{4 x}}

y = \frac{8 Ei ( 4 x )}{e ^{4 x}} + \frac{1 - 8 Ei ( 4 \cdot 0 )}{e ^{4 x}}

\int \frac{1}{( 1 + sin ^{2} ( x ) )} d x

x^{2} - x - 2

\frac{d}{d x} ((10 x + 2)^{x})

x y^{3}

\frac{d y}{d x}, y = x + \frac{1}{x}

What is the solution for y^'x+4yx=8,y(0)=1 ?
The solution for y^'x+4yx=8,y(0)=1 is y=(8\Ei(4x))/(e^{4x)}+(1-8\Ei(4*0))/(e^{4x)}