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

The solution for y^'+y+3cos(x)=0 is y=-(3(sin(x)+cos(x)))/2+(c_{1})/(e^x)

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y^'+y+3cos(x)=0

y^{^{'}} + y + 3 cos (x) = 0

y = - \frac{3 ( sin ( x ) + cos ( x ) )}{2} + \frac{c _{1}}{e ^{x}}

Solution steps

y^{'} (x) + y + 3 cos (x) = 0

Solve linear ODE:

y = - \frac{3 ( sin ( x ) + cos ( x ) )}{2} + \frac{c _{1}}{e ^{x}}

y = - \frac{3 ( sin ( x ) + cos ( x ) )}{2} + \frac{c _{1}}{e ^{x}}

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

x \to 0.9 lim (4 x)

(1 + x^{2}) d y - (1 + 4 y^{2}) d x = 0

\int (x^{2} + 6) \cdot e^{(- x)} d x

\frac{\partial}{\partial z} (3 z e^{x y})

What is the solution for y^'+y+3cos(x)=0 ?
The solution for y^'+y+3cos(x)=0 is y=-(3(sin(x)+cos(x)))/2+(c_{1})/(e^x)