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

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

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y^'+y=(1-x)/(2x^2)

y^{^{'}} + y = \frac{1 - x}{2 x ^{2}}

y = - \frac{1}{2 x} + \frac{c _{1}}{e ^{x}}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{2 x} + \frac{c _{1}}{e ^{x}}

y = - \frac{1}{2 x} + \frac{c _{1}}{e ^{x}}

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x \to - 1 lim (\frac{x}{0})

f (x) = 5 \cdot 3^{x}, at x = 2

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

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