What is the solution for y^'+1/x y=4(x+7)^2 ?

The solution for y^'+1/x y=4(x+7)^2 is y=x^3+(56x^2)/3+98x+(c_{1})/x

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y^'+1/x y=4(x+7)^2

y^{^{'}} + \frac{1}{x} y = 4 (x + 7)^{2}

y = x^{3} + \frac{56 x ^{2}}{3} + 98 x + \frac{c _{1}}{x}

Solution steps

y^{'} (x) + \frac{1}{x} y = 4 (x + 7)^{2}

Solve linear ODE:

y = x^{3} + \frac{56 x ^{2}}{3} + 98 x + \frac{c _{1}}{x}

y = x^{3} + \frac{56 x ^{2}}{3} + 98 x + \frac{c _{1}}{x}

4 x (x^{2} - 1)

\frac{d y}{d t} = 16 - \frac{y}{250 + 0 t}

y = 3 x^{2} + 10 x - 26

\int sin^{5} (x) cos (x) d x

\frac{d}{d x} (x^{s i n (2 x)})

What is the solution for y^'+1/x y=4(x+7)^2 ?
The solution for y^'+1/x y=4(x+7)^2 is y=x^3+(56x^2)/3+98x+(c_{1})/x