What is the solution for y^{''}+4y+8=0,y(0)=4,y^'(0)=-3 ?

The solution for y^{''}+4y+8=0,y(0)=4,y^'(0)=-3 is y=6cos(2t)-3/2 sin(2t)-2

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y^{''}+4y+8=0,y(0)=4,y^'(0)=-3

y^{^{''}} + 4 y + 8 = 0, y (0) = 4, y^{^{'}} (0) = - 3

y = 6 cos (2 t) - \frac{3}{2} sin (2 t) - 2

Solution steps

y^{''} (t) + 4 y + 8 = 0

Solve linear ODE:

y = 6 cos (2 t) - \frac{3}{2} sin (2 t) - 2

y = 6 cos (2 t) - \frac{3}{2} sin (2 t) - 2

f (x) = - 2 \cdot x, at x = 1

\frac{d}{d x} (\frac{6 x ^{2} + 6 x + 2}{x})

\frac{\partial}{\partial x} (cos^{4} (x^{3} y^{8}))

f (x) = arctan (x)

y = x^{4 c o s (x)}

What is the solution for y^{''}+4y+8=0,y(0)=4,y^'(0)=-3 ?
The solution for y^{''}+4y+8=0,y(0)=4,y^'(0)=-3 is y=6cos(2t)-3/2 sin(2t)-2