What is the solution for y^'+10y=2cos(t) ?

The solution for y^'+10y=2cos(t) is y=(2(sin(t)+10cos(t)))/(101)+(c_{1})/(e^{10t)}

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y^'+10y=2cos(t)

y^{^{'}} + 10 y = 2 cos (t)

y = \frac{2 ( sin ( t ) + 10 cos ( t ) )}{101} + \frac{c _{1}}{e ^{10 t}}

Solution steps

y^{'} (t) + 10 y = 2 cos (t)

Solve linear ODE:

y = \frac{2 ( sin ( t ) + 10 cos ( t ) )}{101} + \frac{c _{1}}{e ^{10 t}}

y = \frac{2 ( sin ( t ) + 10 cos ( t ) )}{101} + \frac{c _{1}}{e ^{10 t}}

x y^{^{'}} = 2 y + x^{3} e^{x}

t \to 1 lim (3 t^{3} + 2)

\int \frac{x}{1 + x} d x

\frac{\partial}{\partial x} (- 2 x)

\frac{\partial}{\partial x} (x e^{- 2 x^{2} - 2 y^{2}})

What is the solution for y^'+10y=2cos(t) ?
The solution for y^'+10y=2cos(t) is y=(2(sin(t)+10cos(t)))/(101)+(c_{1})/(e^{10t)}