What is the solution for y^{''}-2y^'+10y=40,y(0)=0,y^'(0)=0 ?

The solution for y^{''}-2y^'+10y=40,y(0)=0,y^'(0)=0 is y=e^t(-4cos(3t)+4/3 sin(3t))+4

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y^{''}-2y^'+10y=40,y(0)=0,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} + 10 y = 40, y (0) = 0, y^{^{'}} (0) = 0

y = e^{t} (- 4 cos (3 t) + \frac{4}{3} sin (3 t)) + 4

Solution steps

y^{''} (t) - 2 y^{'} (t) + 10 y = 40

Solve linear ODE:

y = e^{t} (- 4 cos (3 t) + \frac{4}{3} sin (3 t)) + 4

y = e^{t} (- 4 cos (3 t) + \frac{4}{3} sin (3 t)) + 4

y^{^{''}} - 4 \cdot y = - 3

u^{^{''}} - a \cdot H (x) = x

x^{^{''}} + 5 x^{^{'}} + 4 x = t + 1

y^{^{''}} - 3 y^{^{'}} + 2 y = 3 e^{2 x}

y^{^{''}} + 25 y = 10 cos (5 x)

What is the solution for y^{''}-2y^'+10y=40,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-2y^'+10y=40,y(0)=0,y^'(0)=0 is y=e^t(-4cos(3t)+4/3 sin(3t))+4