What is the solution for y^'=2sin(x)+1+y ?

The solution for y^'=2sin(x)+1+y is y=-cos(x)-sin(x)-1+c_{1}e^x

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y^'=2sin(x)+1+y

y^{^{'}} = 2 sin (x) + 1 + y

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

Solution steps

y^{'} (x) = 2 sin (x) + 1 + y

Solve linear ODE:

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

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

f (x) = x^{3} + 2 x, at x = - 3

n = 1 \sum \infty (\frac{1}{n})^{n}

\int_{0}^{2} (x^{2} - 5 x) d x

y^{^{'}} + y = 5 cos (2 t)

What is the solution for y^'=2sin(x)+1+y ?
The solution for y^'=2sin(x)+1+y is y=-cos(x)-sin(x)-1+c_{1}e^x