Hello,

I have the DE:

\(\displaystyle y''(t) + y = 0\)

with IC's:

\(\displaystyle y'(0) = 1\)hs

\(\displaystyle y(0) = 0\)

If I convert this to a system of 1st order ODES is it just:

\(\displaystyle \frac{dy_1}{dt} = y_2, \ \ \ \frac{dy_2}{dt} = -y_1 \)

How about if the DE is:

\(\displaystyle y''(t) + y = \sin(2t)\) and the same IC's.

Thanks

The first question is a second order homogenious equation. We find the roots by computing the characteristic equation.

\(\displaystyle \lambda^2 + 1 = 0 \)

Clearly this has the imaginary roots \(\displaystyle -i \) and \(\displaystyle i \)

So,

\(\displaystyle y = c_1 e^{-i x} + c_2 e^{ i x } \)

Compute and sub in initial conditions.

The second question can be computed by method of comparing co-efficients (we can also use variation of paramters but co-efficients is easiast here).

For \(\displaystyle y''(t) + y = \sin(2t)\) the general solution is the solution from question 1 plus the particular solution.

For the particular solution,

Let \(\displaystyle y = Asin(2t) + Bcos(2t) \)

So,

\(\displaystyle y^{ \prime } = 2Acos(2t) -2Bsin(2t) \)

\(\displaystyle y^{ \prime \prime } = -4Acos(2t) - 4Bcos(2t) \)

Subbing into the equation we get,

\(\displaystyle -4Acos(2t) - 4Bcos(2t) + Asin(2t) + Bcos(2t) = sin(2t) \)

\(\displaystyle cos(2t) [ - 4A - 3B ] + Asin(2t) = sin(2t) \)

\(\displaystyle A = 1 \)

\(\displaystyle -4 -3B = 0 \)

\(\displaystyle B = - \frac{4}{3} \)

And \(\displaystyle y = sin(2t) - \frac{4}{3}cos(2t) \)