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| | 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: |
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| | <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: |
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| | <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]] |