What is the solution for y^'=e^{-5y}(2sqrt(7x)-4),y(0)=1 ?

The solution for y^'=e^{-5y}(2sqrt(7x)-4),y(0)=1 is y=(ln((20sqrt(7)x^{3/2})/3-20x+e^5))/(5)

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y^'=e^{-5y}(2sqrt(7x)-4),y(0)=1

y^{^{'}} = e^{- 5 y} (2 7 x - 4), y (0) = 1

y = \frac{ln ( \frac{20 7 x ^{\frac{3}{2}}}{3} - 20 x + e ^{5} )}{5}

Solution steps

y^{'} (x) = e^{- 5 y} (2 7 x - 4)

Solve separable ODE:

y = \frac{ln ( \frac{20 7 x ^{\frac{3}{2}}}{3} - 20 x + e ^{5} )}{5}

y = \frac{ln ( \frac{20 7 x ^{\frac{3}{2}}}{3} - 20 x + e ^{5} )}{5}

y^{^{'}} = \frac{y ^{2} + 3}{( 2 x y )}

e^{2 x} \frac{d y}{d x} = \frac{2}{y}

ln (x) y^{^{'}} - (\frac{2 \cdot y}{x}) = 0

\frac{d x}{d y} = sin (5 x)

\frac{d y}{d x} y = ln (ln (ln (x)))

What is the solution for y^'=e^{-5y}(2sqrt(7x)-4),y(0)=1 ?
The solution for y^'=e^{-5y}(2sqrt(7x)-4),y(0)=1 is y=(ln((20sqrt(7)x^{3/2})/3-20x+e^5))/(5)