What is the solution for y^'+3y=te^{4t} ?

The solution for y^'+3y=te^{4t} is y=(e^{4t}(7t-1))/(49)+(c_{1})/(e^{3t)}

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

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

y = \frac{e ^{4 t} ( 7 t - 1 )}{49} + \frac{c _{1}}{e ^{3 t}}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{4 t} ( 7 t - 1 )}{49} + \frac{c _{1}}{e ^{3 t}}

y = \frac{e ^{4 t} ( 7 t - 1 )}{49} + \frac{c _{1}}{e ^{3 t}}

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What is the solution for y^'+3y=te^{4t} ?
The solution for y^'+3y=te^{4t} is y=(e^{4t}(7t-1))/(49)+(c_{1})/(e^{3t)}