What is the solution for 2y^'+ty=7 ?

The solution for 2y^'+ty=7 is y=(7sqrt(pi)\erfi(t/2)}{2e^{(t^2)/4}}+(c_{1})/(e^{\frac{t^2)/4)}

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2y^'+ty=7

2 y^{^{'}} + t y = 7

y = \frac{7 π erfi ( \frac{t}{2} )}{2 e ^{\frac{t ^{2}}{4}}} + \frac{c _{1}}{e ^{\frac{t ^{2}}{4}}}

Solution steps

2 y^{'} (t) + t y = 7

Solve linear ODE:

y = \frac{7 π erfi ( \frac{t}{2} )}{2 e ^{\frac{t ^{2}}{4}}} + \frac{c _{1}}{e ^{\frac{t ^{2}}{4}}}

y = \frac{7 π erfi ( \frac{t}{2} )}{2 e ^{\frac{t ^{2}}{4}}} + \frac{c _{1}}{e ^{\frac{t ^{2}}{4}}}

\frac{d}{d x} (\frac{3 x - 1}{2 x + 5})

y = 4 x^{2}, y^{2} = \frac{1}{4} x

y^{^{''}} - x^{2} + y^{^{'}} = 0

\int \frac{1 - x ^{2}}{x} d x

f (x) = 7^{x}

What is the solution for 2y^'+ty=7 ?
The solution for 2y^'+ty=7 is y=(7sqrt(pi)\erfi(t/2)}{2e^{(t^2)/4}}+(c_{1})/(e^{\frac{t^2)/4)}