What is the solution for y^'=y+1+2sin(t) ?

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

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y^'=y+1+2sin(t)

y^{^{'}} = y + 1 + 2 sin (t)

y = - 1 - cos (t) - sin (t) + c_{1} e^{t}

Solution steps

y^{'} (t) = y + 1 + 2 sin (t)

Solve linear ODE:

y = - 1 - cos (t) - sin (t) + c_{1} e^{t}

y = - 1 - cos (t) - sin (t) + c_{1} e^{t}

x \to 0 + lim (\frac{2 x}{ln ( x )})

\int \frac{cos ( x )}{cos ( x )} d x

8 x^{2} + x y + 8 y^{2} = 17, (1, 1)

y = 4 x^{3} - 12 x + 3

\int x^{2} - 5 x + 2 d x

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