What is the solution for y^{''}+42y=42cos(42t) ?

The solution for y^{''}+42y=42cos(42t) is y=c_{1}cos(sqrt(42)t)+c_{2}sin(sqrt(42)t)-1/41 cos(42t)

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y^{''}+42y=42cos(42t)

y^{^{''}} + 42 y = 42 cos (42 t)

y = c_{1} cos (42 t) + c_{2} sin (42 t) - \frac{1}{41} cos (42 t)

Solution steps

y^{''} (t) + 42 y = 42 cos (42 t)

Solve linear ODE:

y = c_{1} cos (42 t) + c_{2} sin (42 t) - \frac{1}{41} cos (42 t)

y = c_{1} cos (42 t) + c_{2} sin (42 t) - \frac{1}{41} cos (42 t)

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{5 x}, y (0) = 1, y (0) = - 2

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{x} sin (x)

y^{^{''}} + y = \frac{1}{sin ^{2} ( x )}

5 y^{^{''}} + 10 y^{^{'}} = 3 x^{2}

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

What is the solution for y^{''}+42y=42cos(42t) ?
The solution for y^{''}+42y=42cos(42t) is y=c_{1}cos(sqrt(42)t)+c_{2}sin(sqrt(42)t)-1/41 cos(42t)