What is the solution for y^'=(4x+7)(5y+4) ?

The solution for y^'=(4x+7)(5y+4) is y=(e^{10x^2+35x+c_{1}})/5-4/5

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y^'=(4x+7)(5y+4)

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

y = \frac{e ^{10 x^{2} + 35 x + c_{1}}}{5} - \frac{4}{5}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{10 x^{2} + 35 x + c_{1}}}{5} - \frac{4}{5}

y = \frac{e ^{10 x^{2} + 35 x + c_{1}}}{5} - \frac{4}{5}

x^{^{'}} = x^{2} (t)

\frac{d v}{d x} + xv = 0

\frac{d v}{d t} v = (2^{\frac{5}{2}}) t^{\frac{3}{2}} + 3

\frac{d y}{d x} (sin (π) x + cos (π) y)^{2} = 2

x^{2} y - 3 x y^{^{'}} + 4 y = 0

What is the solution for y^'=(4x+7)(5y+4) ?
The solution for y^'=(4x+7)(5y+4) is y=(e^{10x^2+35x+c_{1}})/5-4/5