To solve the second-order differential equation y – 36y” = 0, we start by rewriting the equation in standard form:
y” – rac{1}{36}y = 0
This is a linear homogeneous differential equation with constant coefficients. To solve it, we’ll assume a solution of the form:
y(t) = e^{rt}
where r is a constant that we need to determine. Next, we compute the second derivative:
y”(t) = r^2 e^{rt}
Substituting y(t) and y”(t) into the equation gives:
r^2 e^{rt} – rac{1}{36} e^{rt} = 0
Factoring out e^{rt}, we have:
e^{rt} (r^2 – rac{1}{36}) = 0
Since e^{rt}
eq 0 for any real number r, we can set the characteristic polynomial to zero:
r^2 – rac{1}{36} = 0
Solving for r yields:
r^2 = rac{1}{36}
r = rac{1}{6} ext{ or } r = - rac{1}{6}
With these two distinct roots, we can write the general solution to the differential equation:
y(t) = C_1 e^{ rac{1}{6}t} + C_2 e^{- rac{1}{6}t}
Here, C_1 and C_2 are arbitrary constants determined by initial conditions or boundary values, should they be provided.
In conclusion, the general solution to the differential equation y – 36y” = 0 is:
y(t) = C_1 e^{ rac{1}{6}t} + C_2 e^{- rac{1}{6}t}