What is the solution for xy^'+2y=sin(x),y(pi)= 1/pi ?

The solution for xy^'+2y=sin(x),y(pi)= 1/pi is y=(sin(x)}{x^2}-\frac{cos(x))/x

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xy^'+2y=sin(x),y(pi)= 1/pi

x y^{^{'}} + 2 y = sin (x), y (π) = \frac{1}{π}

y = \frac{sin ( x )}{x ^{2}} - \frac{cos ( x )}{x}

Solution steps

x y^{'} (x) + 2 y = sin (x)

Solve linear ODE:

y = \frac{sin ( x )}{x ^{2}} - \frac{cos ( x )}{x}

y = \frac{sin ( x )}{x ^{2}} - \frac{cos ( x )}{x}

\int \frac{1}{r ^{2} - x ^{2}} d x

\frac{d y}{d x}, x y = 6

\frac{\partial}{\partial y} (x + 2 y^{\frac{1}{2}})

y = \frac{7}{x - 1}

4 cos (5 ln (x))

What is the solution for xy^'+2y=sin(x),y(pi)= 1/pi ?
The solution for xy^'+2y=sin(x),y(pi)= 1/pi is y=(sin(x)}{x^2}-\frac{cos(x))/x