What is the solution for y^{''}-6y^'+34=0 ?

The solution for y^{''}-6y^'+34=0 is y=c_{1}e^{6t}+c_{2}+(17t)/3

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y^{''}-6y^'+34=0

y^{^{''}} - 6 y^{^{'}} + 34 = 0

y = c_{1} e^{6 t} + c_{2} + \frac{17 t}{3}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 34 = 0

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} + \frac{17 t}{3}

y = c_{1} e^{6 t} + c_{2} + \frac{17 t}{3}

y^{^{''}} + 25 y = 5 sec (5 t)

y^{^{''}} + cy - c - k = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{3 x}

y^{^{''}} + y = 3 sin (2 x) + x cos (2 x)

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 159 = 0

What is the solution for y^{''}-6y^'+34=0 ?
The solution for y^{''}-6y^'+34=0 is y=c_{1}e^{6t}+c_{2}+(17t)/3