What is the solution for y^'+y=-cos(6x) ?

The solution for y^'+y=-cos(6x) is y=-(6sin(6x))/(37)-(cos(6x))/(37)+(c_{1})/(e^x)

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y^'+y=-cos(6x)

y^{^{'}} + y = - cos (6 x)

y = - \frac{6 sin ( 6 x )}{37} - \frac{cos ( 6 x )}{37} + \frac{c _{1}}{e ^{x}}

Solution steps

y^{'} (x) + y = - cos (6 x)

Solve linear ODE:

y = - \frac{6 sin ( 6 x )}{37} - \frac{cos ( 6 x )}{37} + \frac{c _{1}}{e ^{x}}

y = - \frac{6 sin ( 6 x )}{37} - \frac{cos ( 6 x )}{37} + \frac{c _{1}}{e ^{x}}

\int x sec^{2} (3 x) d x

o f, x 2 x - 3

y = 9 - x^{2}, y = 0

\frac{\partial}{\partial x} (7 + x ln (x y - 7))

f (x) = θ sin (θ)

What is the solution for y^'+y=-cos(6x) ?
The solution for y^'+y=-cos(6x) is y=-(6sin(6x))/(37)-(cos(6x))/(37)+(c_{1})/(e^x)