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Popular >

y^{''}-2y^'+6=0

Solution

y^{^{''}} - 2 y^{^{'}} + 6 = 0

Solution

y = c_{1} e^{2 t} + c_{2} + 3 t

Solution steps

y^{''} (t) - 2 y^{'} (t) + 6 = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} + 3 t

y = c_{1} e^{2 t} + c_{2} + 3 t

Popular Examples

y^{''}+5y^'+4y=20

y^{^{''}} + 5 y^{^{'}} + 4 y = 20

y^{''}+4y=cos(9/4 t)

y^{^{''}} + 4 y = cos (\frac{9}{4} t)

(d^4y)/(dx^4)+(d^2y)/(dx^2)=cos(4x)

\frac{d ^{4} y}{d x ^{4}} + \frac{d ^{2} y}{d x ^{2}} = cos (4 x)

y^{''''}-3y^{''}-4y=10sin(x)

y^{^{''''}} - 3 y^{^{''}} - 4 y = 10 sin (x)

y^{''}-6y^'+8y=e^{3t},y(0)=0,y^'(0)=0

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{3 t}, y (0) = 0, y^{^{'}} (0) = 0

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-2y^'+6=0 ?
The solution for y^{''}-2y^'+6=0 is y=c_{1}e^{2t}+c_{2}+3t