What is the solution for y^'+2ty=1 ?

The solution for y^'+2ty=1 is y=(sqrt(pi)\erfi(t))/(2e^{t^2)}+(c_{1})/(e^{t^2)}

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y^'+2ty=1

y^{^{'}} + 2 t y = 1

y = \frac{π erfi ( t )}{2 e ^{t^{2}}} + \frac{c _{1}}{e ^{t^{2}}}

Solution steps

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

Solve linear ODE:

y = \frac{π erfi ( t )}{2 e ^{t^{2}}} + \frac{c _{1}}{e ^{t^{2}}}

y = \frac{π erfi ( t )}{2 e ^{t^{2}}} + \frac{c _{1}}{e ^{t^{2}}}

\int_{2}^{\infty} 7 x^{- 2} d x

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

x \to 0 lim (3^{x})

\int x^{3} 7 + x^{4} d x

x^{6}, x^{7}

What is the solution for y^'+2ty=1 ?
The solution for y^'+2ty=1 is y=(sqrt(pi)\erfi(t))/(2e^{t^2)}+(c_{1})/(e^{t^2)}