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

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

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xy^'+2y=-2cos(x)

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

y = - \frac{2 ( x sin ( x ) + cos ( x ) )}{x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

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

Solve linear ODE:

y = - \frac{2 ( x sin ( x ) + cos ( x ) )}{x ^{2}} + \frac{c _{1}}{x ^{2}}

y = - \frac{2 ( x sin ( x ) + cos ( x ) )}{x ^{2}} + \frac{c _{1}}{x ^{2}}

x \to \frac{5 π}{6} lim (\frac{sin ( x )}{cos ( x )})

(2.1) (- 14.7)

640 (1.06)^{t}

f (x) = 2 x^{2} - 32

f (t) = 5 + 4 \cdot δ (t)

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