What is the solution for 4y^'+ty=3 ?

The solution for 4y^'+ty=3 is y=(integral of (3e^{\frac{t^2)/8})/(4)}{e^{(t^2)/8}}

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4y^'+ty=3

4 y^{^{'}} + t y = 3

y = \frac{\int \frac{3 e ^{\frac{t ^{2}}{8}}}{4} d t}{e ^{\frac{t ^{2}}{8}}}

Solution steps

4 y^{'} (t) + t y = 3

Solve linear ODE:

y = \frac{\int \frac{3 e ^{\frac{t ^{2}}{8}}}{4} d t}{e ^{\frac{t ^{2}}{8}}}

y = \frac{\int \frac{3 e ^{\frac{t ^{2}}{8}}}{4} d t}{e ^{\frac{t ^{2}}{8}}}

\int x^{5} e^{2 x} d x

\int csc^{2} (5 x) d x

\int_{1}^{7} (t^{2} - 4 t - 12) d t

x \to 4 - lim (1 + 4 - x)

y = (x^{5} + 2)^{2} (x^{4} + 4)^{4}

What is the solution for 4y^'+ty=3 ?
The solution for 4y^'+ty=3 is y=(integral of (3e^{\frac{t^2)/8})/(4)}{e^{(t^2)/8}}