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

The solution for 2y^'+ty=2 is y=(sqrt(pi)\erfi(t/2)}{e^{(t^2)/4}}+(c_{1})/(e^{\frac{t^2)/4)}

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2y^'+ty=2

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

y = \frac{π erfi ( \frac{t}{2} )}{e ^{\frac{t ^{2}}{4}}} + \frac{c _{1}}{e ^{\frac{t ^{2}}{4}}}

Solution steps

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

Solve linear ODE:

y = \frac{π erfi ( \frac{t}{2} )}{e ^{\frac{t ^{2}}{4}}} + \frac{c _{1}}{e ^{\frac{t ^{2}}{4}}}

y = \frac{π erfi ( \frac{t}{2} )}{e ^{\frac{t ^{2}}{4}}} + \frac{c _{1}}{e ^{\frac{t ^{2}}{4}}}

What is the solution for 2y^'+ty=2 ?
The solution for 2y^'+ty=2 is y=(sqrt(pi)\erfi(t/2)}{e^{(t^2)/4}}+(c_{1})/(e^{\frac{t^2)/4)}