What is the solution for e^yy^'=e^y+7x ?

The solution for e^yy^'=e^y+7x is y=ln(c_{1}+7(-e^{-x}x-e^{-x}))+x

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e^yy^'=e^y+7x

e^{y} y^{^{'}} = e^{y} + 7 x

y = ln (c_{1} + 7 (- e^{- x} x - e^{- x})) + x

Solution steps

e^{y} y^{'} (x) = e^{y} + 7 x

Solve exact ODE:

y = ln (c_{1} + 7 (- e^{- x} x - e^{- x})) + x

y = ln (c_{1} + 7 (- e^{- x} x - e^{- x})) + x

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\frac{d y}{d t} = 0.3 y + 26 e^{0.3 t}

What is the solution for e^yy^'=e^y+7x ?
The solution for e^yy^'=e^y+7x is y=ln(c_{1}+7(-e^{-x}x-e^{-x}))+x