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Popular >

y^{''}+y-12=0

Solution

y^{^{''}} + y - 12 = 0

Solution

y = c_{1} cos (t) + c_{2} sin (t) + 12

Solution steps

y^{''} (t) + y - 12 = 0

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) + 12

y = c_{1} cos (t) + c_{2} sin (t) + 12

Graph

Popular Examples

x^{''}-3x^'+2x=20sin(2t)

x^{^{''}} - 3 x^{^{'}} + 2 x = 20 sin (2 t)

y^{''}-12y^'+36y=8e^{6x}

y^{^{''}} - 12 y^{^{'}} + 36 y = 8 e^{6 x}

x^{''}-5x^'+6x=12,x(0)=2,x^'(0)=0

x^{^{''}} - 5 x^{^{'}} + 6 x = 12, x (0) = 2, x^{^{'}} (0) = 0

y^{''}+2y^'+y=4e^x

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{x}

y^{''}-2y^'+y=e^{t-1},y(1)=0,y^'(1)=5

y^{^{''}} - 2 y^{^{'}} + y = e^{t - 1}, y (1) = 0, y^{^{'}} (1) = 5

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+y-12=0 ?
The solution for y^{''}+y-12=0 is y=c_{1}cos(t)+c_{2}sin(t)+12