What is the solution for y^'=ty-2t-2 ?

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

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y^'=ty-2t-2

y^{^{'}} = t y - 2 t - 2

y = 2 - 2 π erf (\frac{t}{2}) e^{\frac{t ^{2}}{2}} + c_{1} e^{\frac{t ^{2}}{2}}

Solution steps

y^{'} (t) = t y - 2 t - 2

Solve linear ODE:

y = 2 - 2 π erf (\frac{t}{2}) e^{\frac{t ^{2}}{2}} + c_{1} e^{\frac{t ^{2}}{2}}

y = 2 - 2 π erf (\frac{t}{2}) e^{\frac{t ^{2}}{2}} + c_{1} e^{\frac{t ^{2}}{2}}

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

y^{^{''}} + 4 y^{^{'}} + 20 y = - 3 sin (2 x)

\frac{d}{d x} (\frac{ln ( x ^{2} + 1 )}{3 x + 1})

1 + ln (x) + \frac{y}{x} d x = (8 - ln (x)) d y

\frac{d}{d x} (x sin (x + y))

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