What is the solution for y^'+y=cos(t),y(0)=1 ?

The solution for y^'+y=cos(t),y(0)=1 is y=(sin(t))/2+(cos(t))/2+1/(2e^t)

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y^'+y=cos(t),y(0)=1

y^{^{'}} + y = cos (t), y (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

y = lo g_{e} (lo g_{e} (x))

h \to 0 lim (\frac{( h + 2 ) ^{3} - 8}{h})

\int \frac{1}{y} d y

\int_{0}^{1} 5 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

(\frac{8 x + 5}{8 x - 5})^{\frac{1}{2}}

What is the solution for y^'+y=cos(t),y(0)=1 ?
The solution for y^'+y=cos(t),y(0)=1 is y=(sin(t))/2+(cos(t))/2+1/(2e^t)