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

The solution for y^'=7e^{4x-5y},y(0)=0 is y=(ln(35/4 e^{4x}-31/4))/5

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

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

y = \frac{ln ( \frac{35}{4} e ^{4 x} - \frac{31}{4} )}{5}

Solution steps

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

Solve separable ODE:

y = \frac{ln ( \frac{35}{4} e ^{4 x} - \frac{31}{4} )}{5}

y = \frac{ln ( \frac{35}{4} e ^{4 x} - \frac{31}{4} )}{5}

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What is the solution for y^'=7e^{4x-5y},y(0)=0 ?
The solution for y^'=7e^{4x-5y},y(0)=0 is y=(ln(35/4 e^{4x}-31/4))/5