What is the solution for y^'= y/2+4e^{t/2} ?

The solution for y^'= y/2+4e^{t/2} is y=4e^{t/2}t+c_{1}e^{t/2}

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y^'= y/2+4e^{t/2}

y^{^{'}} = \frac{y}{2} + 4 e^{\frac{t}{2}}

y = 4 e^{\frac{t}{2}} t + c_{1} e^{\frac{t}{2}}

Solution steps

y^{'} (t) = \frac{y}{2} + 4 e^{\frac{t}{2}}

Solve linear ODE:

y = 4 e^{\frac{t}{2}} t + c_{1} e^{\frac{t}{2}}

y = 4 e^{\frac{t}{2}} t + c_{1} e^{\frac{t}{2}}

\frac{d}{d x} (\frac{x}{( x ^{2} - 1 )})

\frac{\partial}{\partial y} (ln (\frac{x}{x ^{2} + y ^{2}}))

\int t^{3} 1 + t^{2} d t

\int x 4^{x} d x

y = (cos (x))^{s i n (x)}

What is the solution for y^'= y/2+4e^{t/2} ?
The solution for y^'= y/2+4e^{t/2} is y=4e^{t/2}t+c_{1}e^{t/2}