What is the solution for y^'-4y=10sin(2t),y(0)=0 ?

The solution for y^'-4y=10sin(2t),y(0)=0 is y=-cos(2t)-2sin(2t)+e^{4t}

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y^'-4y=10sin(2t),y(0)=0

y^{^{'}} - 4 y = 10 sin (2 t), y (0) = 0

y = - cos (2 t) - 2 sin (2 t) + e^{4 t}

Solution steps

y^{'} (t) - 4 y = 10 sin (2 t)

Solve linear ODE:

y = - cos (2 t) - 2 sin (2 t) + e^{4 t}

y = - cos (2 t) - 2 sin (2 t) + e^{4 t}

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

\int \frac{x}{2 x ^{2} + 3} d x

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

\int \frac{x ^{2} + 3 x + 7}{x} d x

d er i v a t i v e y = 5 x^{2} e^{2 x}

What is the solution for y^'-4y=10sin(2t),y(0)=0 ?
The solution for y^'-4y=10sin(2t),y(0)=0 is y=-cos(2t)-2sin(2t)+e^{4t}