What is the solution for y^'+y=e^{-4t}cos(3t),y(0)=0 ?

The solution for y^'+y=e^{-4t}cos(3t),y(0)=0 is y=(sin(3t))/(6e^{4t)}-(cos(3t))/(6e^{4t)}+1/(6e^t)

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y^'+y=e^{-4t}cos(3t),y(0)=0

y^{^{'}} + y = e^{- 4 t} cos (3 t), y (0) = 0

y = \frac{sin ( 3 t )}{6 e ^{4 t}} - \frac{cos ( 3 t )}{6 e ^{4 t}} + \frac{1}{6 e ^{t}}

Solution steps

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

Solve linear ODE:

y = \frac{sin ( 3 t )}{6 e ^{4 t}} - \frac{cos ( 3 t )}{6 e ^{4 t}} + \frac{1}{6 e ^{t}}

y = \frac{sin ( 3 t )}{6 e ^{4 t}} - \frac{cos ( 3 t )}{6 e ^{4 t}} + \frac{1}{6 e ^{t}}

x^{2} y^{^{'}} + x y = 3

e^{- 2 t + 3}

\int \frac{4}{θ ^{2}} cos (\frac{4}{θ}) d θ

\int \frac{1}{x ( x - 4 )} d x

f (x) = \frac{2 t}{2 t}

What is the solution for y^'+y=e^{-4t}cos(3t),y(0)=0 ?
The solution for y^'+y=e^{-4t}cos(3t),y(0)=0 is y=(sin(3t))/(6e^{4t)}-(cos(3t))/(6e^{4t)}+1/(6e^t)