Download original notes
PAGE 1

3.6 Variation of Parameters

Another way to solve the nonhomogeneous equation:

\[ y'' + p(t)y' + q(t)y = g(t) \]

Solution: \( y = c_1 y_1 + c_2 y_2 + Y \)

  • Complementary: Solution to \( y'' + p(t)y' + q(t)y = 0 \)
  • Particular: \( Y \)

Undetermined Coefficients

  • \( Y \) resembles \( g(t) \)
  • \( Y \) keeps its form after differentiation
\[ y'' + y' = \sec(t) \]

Can't handle this with undetermined coefficients!

\( Y = ? \)

If we tried \( Y = A \sec(t) \):

\[ Y' = A \sec(t) \tan(t) \]

Changed form!

What if \( Y = A \sec(t) + B \tan(t) \)?

\[ Y' = A \sec(t) \tan(t) + B \sec^2(t) \]
PAGE 2

Variation of Parameters: Solution is of the form

\[ y = u_1(t)y_1 + u_2(t)y_2 \]

"parameters" (\( u_1, u_2 \))

complementary (\( y_1, y_2 \))

If we can find \( u_1(t) \) and \( u_2(t) \), we can separate out the particular solution \( Y \).

\[ y'' + p(t)y' + q(t)y = g(t) \]

Solution: \( y = u_1(t)y_1 + u_2(t)y_2 \) (sub into the equation above)

\[ y' = u_1 y_1' + u_1' y_1 + u_2 y_2' + u_2' y_2 \]\[ y'' = u_1 y_1'' + u_1' y_1' + u_1' y_1' + u_1'' y_1 + u_2 y_2'' + u_2' y_2' + u_2' y_2' + u_2'' y_2 \]

If we sub these into \( y'' + py' + qy = g \), eventually we will have TWO 2nd-order eqs. for \( u_1 \) and \( u_2 \).

Two unknowns (\( u_1, u_2 \)), one eq: \( y'' + py' + qy = g \)

→ we can impose one condition

PAGE 3

So, we will impose the condition that eliminates the 2nd-order eqs. (so just \( u_1' \) and \( u_2' \))

\[ u_1' y_1 + u_2' y_2 = 0 \]

Update the eqs:

\[ y = u_1 y_1 + u_2 y_2 \]

\[ y' = u_1 y_1' + u_2 y_2' \]

\[ y'' = u_1 y_1'' + u_1' y_1' + u_2 y_2'' + u_2' y_2' \]

Sub into \[ y'' + py' + qy = g \]

\[ u_1 y_1'' + u_1' y_1' + u_2 y_2'' + u_2' y_2' + p u_1 y_1' + p u_2 y_2' + q u_1 y_1 + q u_2 y_2 = g \]

\[ u_1 (y_1'' + p y_1' + q y_1) + u_2 (y_2'' + p y_2' + q y_2) + u_1' y_1' + u_2' y_2' = g \]

0

because \( y_1 \) is a complementary solution \( (y'' + py' + qy = 0) \)

0

same reason

so,

\[ u_1' y_1' + u_2' y_2' = g \]

solve the two boxed eqs for \( u_1' \) and \( u_2' \)

then integrate to find \( u_1, u_2 \)

PAGE 4

Example

\[ y'' + y = \sec(t) \]

Complementary: \( y'' + y = 0 \)    \( r^2 + 1 = 0 \)    \( r = \pm i \)

\[ y_1 = \cos(t) \]

\[ y_2 = \sin(t) \]

\[ y = u_1 y_1 + u_2 y_2 \]

solve:

\[ u_1' y_1 + u_2' y_2 = 0 \]

\[ u_1' y_1' + u_2' y_2' = g(t) = \sec(t) \]

\[ u_1' \cos(t) + u_2' \sin(t) = 0 \]

\[ -u_1' \sin(t) + u_2' \cos(t) = \sec(t) \]

\[ u_1' = -\frac{\sin(t)}{\cos(t)} u_2' \]

Sub into 2nd eq.

\[ u_2' \frac{\sin^2(t)}{\cos(t)} + u_2' \cos(t) = \sec(t) \]

mult. by \( \cos(t) \)

\[ u_2' [\sin^2(t) + \cos^2(t)] = 1 \]

PAGE 5
\[ u_2' = 1 \rightarrow \text{ } \boxed{u_2 = t + C_2} \]
\[ u_1' = - \frac{\sin(t)}{\cos(t)} u_2' \]
\[ u_1' = - \frac{\sin(t)}{\cos(t)} \rightarrow \text{ } \boxed{u_1 = \ln |\cos(t)| + C_1} \]

General solution to \( y'' + y = \sec(t) \) is then

\[ y = u_1 y_1 + u_2 y_2 \]
\[ = (\ln |\cos(t)| + C_1) \cos(t) + (t + C_2) \sin(t) \]
\[ = \underbrace{C_1 \cos(t) + C_2 \sin(t)}_{\text{complementary}} + \underbrace{\cos(t) \ln |\cos(t)| + t \sin(t)}_{\text{particular (Y)}} \]
PAGE 6
\[ u_1' y_1 + u_2' y_2 = 0 \]
\[ u_1' y_1' + u_2' y_2' = g \]
\[ \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} \begin{bmatrix} u_1' \\ u_2' \end{bmatrix} = \begin{bmatrix} 0 \\ g \end{bmatrix} \]
\[ \begin{bmatrix} u_1' \\ u_2' \end{bmatrix} = \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix}^{-1} \begin{bmatrix} 0 \\ g \end{bmatrix} \]
\[ = \frac{1}{\underbrace{\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}}_{\text{Wronskian}}} \begin{bmatrix} y_2' & -y_2 \\ -y_1' & y_1 \end{bmatrix} \begin{bmatrix} 0 \\ g \end{bmatrix} \]
\[ \begin{bmatrix} u_1' \\ u_2' \end{bmatrix} = \frac{1}{W} \begin{bmatrix} -y_2 g \\ y_1 g \end{bmatrix} \]
PAGE 7
\[ u_1' = - \frac{y_2 g}{W} \]
\[ u_2' = \frac{y_1 g}{W} \]

undetermined coeff: multiply by \( t \) if partial right side copies \( y_1 \) or \( y_2 \)

variation of parameters takes care of that automatically