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Solving Second-Order Non- Homogeneous Differential Equations II Peter Symonds Solving Second-Order Non-Homogeneous Differential Equations II Peter Symonds University of Manchester CHEN20041, Semester 1, 2016
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Solving Second-Order Non- Homogeneous Differential Equations II Peter Symonds Problems with f ( x ) Consider the differential equation d 2 y dx 2 + 3 dy dx + 2 y = 12 e - x . The homogeneous solution is y = Pe - x + Qe - 2 x . For the particular integral, try y PI = ae - x , so dy dx = - ae - x and d 2 y dx 2 = ae - x .
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Solving Second-Order Non- Homogeneous Differential Equations II Peter Symonds Problems with f ( x ) Substituting back into the original equation, we get ae - x - 3 ae - x + 2 ae - x = 12 e - x , and so 0 = 12 e - x . However, there is no value of a that satisfies this equation. The particular integral is not of the form y PI = ae - x .
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Solving Second-Order Non- Homogeneous Differential Equations II Peter Symonds Problems with f ( x ) Instead, for the particular integral try y PI = axe - x . So dy dx = ae - x - axe - x and d 2 y dx 2 = - 2 ax - x + axe - x . Then we get ( - 2 ae - x + axe - x ) + 3( ae - x - axe - x ) + 2 axe - x = 12 e - x . The terms in xe - x cancel, and we have - 2 ae - x + 3 ae - x = 12 e - x , which implies a = 12.
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Solving Second-Order Non- Homogeneous Differential Equations II Peter Symonds Problems with f ( x ) So the particular integral is y PI = 12 xe - x , and the general solution is y = Pe - x + Qe - 2 x + 12 xe - x .
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Solving Second-Order Non- Homogeneous Differential Equations II Peter Symonds Problems with f ( x ) The problem was that original choice of particular integral, y PI = ae - x , was of the same form as part of the homogeneous solution y = Pe - x + Qe - 2 x , so when we substituted into our ODE we got zero. To get around this we multiplied the form for the particular integral by x , to get y PI = axe - x .
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Solving Second-Order Non- Homogeneous Differential Equations II Peter Symonds Problems with f ( x ) Therefore, if the obvious choice for the form of the particular integral is of the same form as (part of) the homogeneous solution, we should multiply this choice of integral by x , and then try to find the values of the constants.
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Solving Second-Order Non- Homogeneous Differential Equations II Peter Symonds
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