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}

f (x) = - \frac{3}{e ^{3 x}}

x \to - 3 lim (x^{2} + 4 x + 1)

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

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

\int X d 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}