What is the solution for y^'+6y=e^{4t},y(0)=2 ?

The solution for y^'+6y=e^{4t},y(0)=2 is y=(e^{4t})/(10)+(19)/(10e^{6t)}

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y^'+6y=e^{4t},y(0)=2

y^{^{'}} + 6 y = e^{4 t}, y (0) = 2

y = \frac{e ^{4 t}}{10} + \frac{19}{10 e ^{6 t}}

Solution steps

y^{'} (t) + 6 y = e^{4 t}

Solve linear ODE:

y = \frac{e ^{4 t}}{10} + \frac{19}{10 e ^{6 t}}

y = \frac{e ^{4 t}}{10} + \frac{19}{10 e ^{6 t}}

\int \frac{x ^{2}}{20 - 4 x + x ^{2}} d x

\frac{d}{d x} (0.5 sin^{2} (x) - x + 1)

\frac{4}{( s + 1 ) ^{2}}

t \cdot y^{^{'}} + t y = 1 - y

t^{2} \cdot e^{a \cdot t}

What is the solution for y^'+6y=e^{4t},y(0)=2 ?
The solution for y^'+6y=e^{4t},y(0)=2 is y=(e^{4t})/(10)+(19)/(10e^{6t)}