What is the solution for e^yy^'=e^y+5x,y(3)=7 ?

The solution for e^yy^'=e^y+5x,y(3)=7 is y=ln(e^4+(20)/(e^3)+5(-e^{-x}x-e^{-x}))+x

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e^yy^'=e^y+5x,y(3)=7

e^{y} y^{^{'}} = e^{y} + 5 x, y (3) = 7

y = ln (e^{4} + \frac{20}{e ^{3}} + 5 (- e^{- x} x - e^{- x})) + x

Solution steps

e^{y} y^{'} (x) = e^{y} + 5 x

Solve exact ODE:

y = ln (e^{4} + \frac{20}{e ^{3}} + 5 (- e^{- x} x - e^{- x})) + x

y = ln (e^{4} + \frac{20}{e ^{3}} + 5 (- e^{- x} x - e^{- x})) + x

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What is the solution for e^yy^'=e^y+5x,y(3)=7 ?
The solution for e^yy^'=e^y+5x,y(3)=7 is y=ln(e^4+(20)/(e^3)+5(-e^{-x}x-e^{-x}))+x