What is the solution for y^'+4x^{-1}y=x^2,\quad y(1)=-2 ?

The solution for y^'+4x^{-1}y=x^2,\quad y(1)=-2 is y=(x^3)/7-(15)/(7x^4)

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y^'+4x^{-1}y=x^2,\quad y(1)=-2

y^{^{'}} + 4 x^{- 1} y = x^{2}, y (1) = - 2

y = \frac{x ^{3}}{7} - \frac{15}{7 x ^{4}}

Solution steps

y^{'} (x) + 4 x^{- 1} y = x^{2}

Solve linear ODE:

y = \frac{x ^{3}}{7} - \frac{15}{7 x ^{4}}

y = \frac{x ^{3}}{7} - \frac{15}{7 x ^{4}}

n = 1 \sum \infty \frac{5 ^{n}}{n ^{5}}

\frac{1}{x + 2}, 0

k = 2 \sum \infty \frac{1}{k} - \frac{1}{4 ^{k}}

(1 + ln (x) + \frac{y}{x}) d x - (1 - ln (x)) d y = 0

\frac{d y}{d x} = 2 x + y

What is the solution for y^'+4x^{-1}y=x^2,\quad y(1)=-2 ?
The solution for y^'+4x^{-1}y=x^2,\quad y(1)=-2 is y=(x^3)/7-(15)/(7x^4)