Difference between revisions of "Reduction of order"
(Created page with ''''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 on…') |
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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>.) |
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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>. | ||
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2. We apply the product rule to find: | 2. We apply the product rule to find: | ||
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<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> | ||
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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> | ||
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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> | ||
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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: | ||
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<math>v(t) = c_1 t + c_2</math> | <math>v(t) = c_1 t + c_2</math> | ||
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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> | ||
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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]] |
Latest revision as of 20:30, January 1, 2026
Reduction of order is a process used to find the solution of a differential equation , when Euler substitution methods find only one value for . (That is, and .)
The process is carried out in the following manner:
1. . However, we must find . cannot be a multiple of , so we assume that .
2. We apply the product rule to find:
3. We substitute these expressions into the initial differential equation:
4. We collect the , , and terms:
5. The terms and equal zero in most instances, leading to the conclusion:
Integrating twice, we yield:
6. This is enough to say that
7. The solution is then
