What is the solution for y^'+y=cos(t) ?

The solution for y^'+y=cos(t) is y=(sin(t))/2+(cos(t))/2+(c_{1})/(e^t)

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y^'+y=cos(t)

y^{^{'}} + y = cos (t)

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

Solution steps

y^{'} (t) + y = cos (t)

Solve linear ODE:

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

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

x \to - \infty lim (3)

\frac{d}{d x} (3^{s i n (3 x)})

\int \frac{x}{4 + x} d x

e (t + 1)

\frac{\partial}{\partial y} (3 x^{4} e^{y})

What is the solution for y^'+y=cos(t) ?
The solution for y^'+y=cos(t) is y=(sin(t))/2+(cos(t))/2+(c_{1})/(e^t)