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'''Reduction of order''' is a process used to find the solution of a differential equation <math>y'' + p(t)y' + q(t)y = 0</math>, when [[Euler substitution]] methods find only one value for <math>\lambda</math>. (That is, <math>\sqrt {b^2-4ac} = 0</math> and <math>\lambda = \frac{-b}{2a}</math>.)

'''Reduction of order''' is a process used to find the solution of a [[differential equation]] <math>y'' + p(t)y' + q(t)y = 0</math>, when [[Euler substitution]] methods find only one value for <math>\lambda</math>. (That is, <math>\sqrt {b^2-4ac} = 0</math> and <math>\lambda = \frac{-b}{2a}</math>.)

 

The process is carried out in the following manner:

The process is carried out in the following manner:

1. <math>y_1 = e^{\lambda t}</math>. However, we must find <math>y_2</math>. <math>y_2</math> cannot be a multiple of <math>y_1</math>, so we assume that <math>y_2 = v(t) y_1</math>.

1. <math>y_1 = e^{\lambda t}</math>. However, we must find <math>y_2</math>. <math>y_2</math> cannot be a multiple of <math>y_1</math>, so we assume that <math>y_2 = v(t) y_1</math>.

2. We apply the product rule to find:

2. We apply the product rule to find:

<math>y_2'' = v(t)'' y_1 + 2 v(t)' y_1' + v(t) y_1''</math>

<math>y_2'' = v(t)'' y_1 + 2 v(t)' y_1' + v(t) y_1''</math>

3. We substitute these expressions into the initial differential equation:

3. We substitute these expressions into the initial differential equation:

<math>y_2'' + p(t)y_2' + q(t)y = (v(t)'' y_1 + 2 v(t)' y_1' + v(t) y_1'') + p(t)(v(t)' y_1 + v(t) y_1') + q(t)(v(t) y_1)</math>

<math>y_2'' + p(t)y_2' + q(t)y = (v(t)'' y_1 + 2 v(t)' y_1' + v(t) y_1'') + p(t)(v(t)' y_1 + v(t) y_1') + q(t)(v(t) y_1)</math>

4. We collect the <math>v(t)''</math>, <math>v(t)'</math>, and <math>v(t)</math> terms:

4. We collect the <math>v(t)''</math>, <math>v(t)'</math>, and <math>v(t)</math> terms:

<math>v(t)''(y_1) + v(t)'(2y_1' + p(t) y_1) + v(t)(y_1'' + p(t) y_1' + q(t) v(t) y_1)</math>

<math>v(t)''(y_1) + v(t)'(2y_1' + p(t) y_1) + v(t)(y_1'' + p(t) y_1' + q(t) v(t) y_1)</math>

5. The terms <math>(2y_1' + p(t) y_1)</math> and <math>(y_1'' + p(t) y_1' + q(t) v(t) y_1)</math> equal zero in most instances, leading to the conclusion:

5. The terms <math>(2y_1' + p(t) y_1)</math> and <math>(y_1'' + p(t) y_1' + q(t) v(t) y_1)</math> equal zero in most instances, leading to the conclusion:

<math>v(t) = c_1 t + c_2</math>

<math>v(t) = c_1 t + c_2</math>

6. This is enough to say that <math>y_2 = v(t) y_1 = ty_1 = t e^{\lambda t}</math>

6. This is enough to say that <math>y_2 = v(t) y_1 = ty_1 = t e^{\lambda t}</math>

7. The solution is then <math>y = c_1 y_1 + c_2 y_2 = e^{\lambda t} (c_1 + c_2 t)</math>

7. The solution is then <math>y = c_1 y_1 + c_2 y_2 = e^{\lambda t} (c_1 + c_2 t)</math>

[[Category: Differential Equations]]

[[Category:Differential Equations]]