What is the solution for y^'=(e^{4x+y}-sec(7x)e^y) ?

The solution for y^'=(e^{4x+y}-sec(7x)e^y) is y=-ln(-1/4 e^{4x}+1/7 ln(tan(7x)+sec(7x))-c_{1})

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

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

y = - ln (- \frac{1}{4} e^{4 x} + \frac{1}{7} ln (tan (7 x) + sec (7 x)) - c_{1})

Solution steps

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

Solve separable ODE:

y = - ln (- \frac{1}{4} e^{4 x} + \frac{1}{7} ln (tan (7 x) + sec (7 x)) - c_{1})

y = - ln (- \frac{1}{4} e^{4 x} + \frac{1}{7} ln (tan (7 x) + sec (7 x)) - c_{1})

3 y^{^{'}} + 4 y = 2

y^{^{'}} + (\frac{1 + y ^{2}}{y}) sin (x) = 0, y (0) = 2

y + x \cdot \frac{d y}{d x} = 0

(x^{2} - 3) y^{- 1} + 2 x y^{^{'}} = 0

y^{^{'}} = y (1 - x), y (1) = e

What is the solution for y^'=(e^{4x+y}-sec(7x)e^y) ?
The solution for y^'=(e^{4x+y}-sec(7x)e^y) is y=-ln(-1/4 e^{4x}+1/7 ln(tan(7x)+sec(7x))-c_{1})