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

The solution for (7+x)y^'+y=7+x is y=(7x)/(7+x)+(x^2)/(2(7+x))+(c_{1})/(7+x)

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

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

y = \frac{7 x}{7 + x} + \frac{x ^{2}}{2 ( 7 + x )} + \frac{c _{1}}{7 + x}

Solution steps

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

Solve linear ODE:

y = \frac{7 x}{7 + x} + \frac{x ^{2}}{2 ( 7 + x )} + \frac{c _{1}}{7 + x}

y = \frac{7 x}{7 + x} + \frac{x ^{2}}{2 ( 7 + x )} + \frac{c _{1}}{7 + x}

sec^{2} (x) - tan^{2} (x)

\frac{\partial}{\partial x} (\frac{x ^{4}}{6 y} + \frac{12 y ^{4}}{x})

\frac{d}{d x} (\frac{x e ^{x}}{x + e ^{x}})

\frac{d}{d x} (sin (x) + \frac{1}{x})

\frac{\partial}{\partial x} (- 3 sin (x))

What is the solution for (7+x)y^'+y=7+x ?
The solution for (7+x)y^'+y=7+x is y=(7x)/(7+x)+(x^2)/(2(7+x))+(c_{1})/(7+x)