What is the solution for y^'=t(1+y/4),y(1)=2.4 ?

The solution for y^'=t(1+y/4),y(1)=2.4 is y=(32e^{(t^2-1)/8})/(5)-4

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y^'=t(1+y/4),y(1)=2.4

y^{^{'}} = t (1 + \frac{y}{4}), y (1) = 2.4

y = \frac{32 e ^{\frac{t ^{2} - 1}{8}}}{5} - 4

Solution steps

y^{'} (t) = t (1 + \frac{y}{4})

Solve separable ODE:

y = \frac{32 e ^{\frac{t ^{2} - 1}{8}}}{5} - 4

y = \frac{32 e ^{\frac{t ^{2} - 1}{8}}}{5} - 4

\frac{d y}{d x} = \frac{5 x}{e ^{y}}

\frac{d v}{d t} = - 2 v

\frac{x ^{2} - 9}{x} \frac{d y}{d x} + y = 0

(x y + x) d x = (x^{2} y^{2} + x^{2} + y^{2} + 1) d y, y (1) = 2

\frac{d y}{d x} = - \frac{3}{7} x^{9} y^{8}

What is the solution for y^'=t(1+y/4),y(1)=2.4 ?
The solution for y^'=t(1+y/4),y(1)=2.4 is y=(32e^{(t^2-1)/8})/(5)-4