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

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

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y^'-cos(t)y=te^{t^2+sin(t)}

y^{^{'}} - cos (t) y = t e^{t^{2} + s i n (t)}

y = \frac{e ^{t^{2} + s i n (t)}}{2} + c_{1} e^{s i n (t)}

Solution steps

y^{'} (t) - cos (t) y = t e^{t^{2} + s i n (t)}

Solve linear ODE:

y = \frac{e ^{t^{2} + s i n (t)}}{2} + c_{1} e^{s i n (t)}

y = \frac{e ^{t^{2} + s i n (t)}}{2} + c_{1} e^{s i n (t)}

\frac{d}{d x} (20 (1 - e^{- \frac{x}{2}}))

\int - 3 x^{2} + 100 d x

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

4 x y \frac{d y}{d x} = 3

\frac{d}{d x} ((x - 1)^{2} e^{x})

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