What is the solution for (7+x)y^'=(x+8)(y-7) ?

The solution for (7+x)y^'=(x+8)(y-7) is y=7e^{x+c_{1}}+e^{x+c_{1}}x+7

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(7+x)y^'=(x+8)(y-7)

(7 + x) y^{^{'}} = (x + 8) (y - 7)

y = 7 e^{x + c_{1}} + e^{x + c_{1}} x + 7

Solution steps

(7 + x) y^{'} (x) = (x + 8) (y - 7)

Solve separable ODE:

y = 7 e^{x + c_{1}} + e^{x + c_{1}} x + 7

y = 7 e^{x + c_{1}} + e^{x + c_{1}} x + 7

\frac{d y}{d x} = e^{x + y}, e^{x} + e^{y} = c

2 x \frac{d y}{d x} - 6 y = 0

2 d x - e^{y - x} d y = 0

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

\frac{d S}{d t} = 0.06 - \frac{3 S}{1420}, t = 10

What is the solution for (7+x)y^'=(x+8)(y-7) ?
The solution for (7+x)y^'=(x+8)(y-7) is y=7e^{x+c_{1}}+e^{x+c_{1}}x+7