What is the solution for xy^'+y=3cos(x),y(pi)=0 ?

The solution for xy^'+y=3cos(x),y(pi)=0 is y=(3sin(x))/x

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xy^'+y=3cos(x),y(pi)=0

x y^{^{'}} + y = 3 cos (x), y (π) = 0

y = \frac{3 sin ( x )}{x}

Solution steps

x y^{'} (x) + y = 3 cos (x)

Solve linear ODE:

y = \frac{3 sin ( x )}{x}

y = \frac{3 sin ( x )}{x}

\frac{\partial}{\partial x} (e^{x} - x (sin (y)))

\frac{6 x ^{2} + 4 x + 2}{x}

\frac{d}{d x} (- x - 1)

x \to + 1 lim (\frac{1 - x}{x ^{2} - 1})

\int_{0}^{1} x e^{- 6 x^{2}} d x

What is the solution for xy^'+y=3cos(x),y(pi)=0 ?
The solution for xy^'+y=3cos(x),y(pi)=0 is y=(3sin(x))/x