MATH 20D Lecture 7

Linear Second-Order ODEs

Second-Order Linear ODE

A second-order linear ODE has the general form

In our case, we consider only second-order ODE with constant coefficients:

In the special case where , we obtain the homogenous form

For the IVP, need two initial conditions: and

Linear Independence of Two Functions

Two functions and are said to be linearly independent on the interval if and only if neither of them is a constant multiple of the other, for all .

A condition for Linear Independence of Solutions

Two solutions and are linearly independent on the interval if

y_{1}‘(t) & y_{2}‘(t)\end{vmatrix}=y_{1}(t)y_{2}‘(t)-y_{1}‘(t)y_{2}(t)\neq 0$$

for some value . If for all then and are linearly dependent.
Here, is called the Wronskian of .

General solution of Homogeneous second-order ODE

In general, if and are two linearly independent solutions of the homogeneous second-order ODE then the general solution to this ODE is given by

for constants .
In the case the characteristic equation has repeated roots then the general solution to the corresponding homogeneous 2nd-order ODE is given by

for constants .

Characteristic Equation with Complex Roots

$y^{\prime \prime }+k^{2}y=0,y(0)=y_0,y^{\prime }(0)=y_1$

Show that a solution of this equation has the general form



So is indeed a soln to
Can you do the same for ?
To show are linearly independent we calculate their Wronskian

-k\sin(kt) & k\cos(kt)\end{vmatrix}=(\cos(kt))(k\cos(kt))-(-k\sin(kt))(\sin(kt))=k\cos^2(kt)+k\sin^2(kt)=k\neq 0$$

Therefore, will be our general soln to

Now let’s solve the same problem using its characteristic equation

$y^{\prime \prime }+k^{2}y=0,y(0)=y_0,y^{\prime }(0)=y_1$

Chara. Eqn.:
So according to Case 1, we do have 2 distinct soln. from the chara. eqn.

Euler's Formula

If
If