What is the solution for y^'=-5sin(3t)+2cos(4t) ?

The solution for y^'=-5sin(3t)+2cos(4t) is y= 5/3 cos(3t)+1/2 sin(4t)+c_{1}

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y^'=-5sin(3t)+2cos(4t)

y^{^{'}} = - 5 sin (3 t) + 2 cos (4 t)

y = \frac{5}{3} cos (3 t) + \frac{1}{2} sin (4 t) + c_{1}

Solution steps

y^{'} (t) = - 5 sin (3 t) + 2 cos (4 t)

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int - 5 sin (3 t) + 2 cos (4 t) d t

\int - 5 sin (3 t) + 2 cos (4 t) d t = \frac{5}{3} cos (3 t) + \frac{1}{2} sin (4 t) + c_{1}

y = \frac{5}{3} cos (3 t) + \frac{1}{2} sin (4 t) + c_{1}

x \to 0 lim (\frac{1}{sin ( 5 x )})

a re a y = - x^{2} + 16, x = - 2, y = 0, x = 3

\int e^{2 - 3 x} d x

x \to (0) + lim (3 x^{(\frac{x}{2})})

\frac{d}{d x} (14 x e^{x})

What is the solution for y^'=-5sin(3t)+2cos(4t) ?
The solution for y^'=-5sin(3t)+2cos(4t) is y= 5/3 cos(3t)+1/2 sin(4t)+c_{1}