What is the solution for y^'-8y2e-5x=0,y(0)=7 ?

The solution for y^'-8y2e-5x=0,y(0)=7 is y=-(5(16ex+1))/(256e^2)+(7+5/(256e^2))e^{16ex}

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y^'-8y2e-5x=0,y(0)=7

y^{^{'}} - 8 y 2 e - 5 x = 0, y (0) = 7

y = - \frac{5 ( 16 e x + 1 )}{256 e ^{2}} + (7 + \frac{5}{256 e ^{2}}) e^{16 e x}

Solution steps

y^{'} (x) - 8 y 2 e - 5 x = 0

Solve linear ODE:

y = - \frac{5 ( 16 e x + 1 )}{256 e ^{2}} + (7 + \frac{5}{256 e ^{2}}) e^{16 e x}

y = - \frac{5 ( 16 e x + 1 )}{256 e ^{2}} + (7 + \frac{5}{256 e ^{2}}) e^{16 e x}

\frac{5}{2 s + 2}

\frac{\partial}{\partial x} (4 e^{2 x})

\frac{x}{x ^{3} + 1}

\frac{d y}{d x} = 3 x cos (x^{2})

\int \frac{3}{5 - 9 x ^{2}} d x

What is the solution for y^'-8y2e-5x=0,y(0)=7 ?
The solution for y^'-8y2e-5x=0,y(0)=7 is y=-(5(16ex+1))/(256e^2)+(7+5/(256e^2))e^{16ex}