inverse of an operator in differential equation
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We have learned in class how to use inverse operator methods to solve ODE's (i.e. with the symbolic DD). E.g,
If I were asked to find a particular solution, ypyp to (D−1)(D−2)[y]=x2ex(D−1)(D−2)[y]=x2ex, then I would use the formula g(x)D−a=eax∫e−axg(x)dxg(x)D−a=eax∫e−axg(x)dx, where g(x)=x2exg(x)=x2ex.
However, here's the difficulty I'm facing; all the ODE's that we have learned in class with operator methods have been dealing with constant coefficient ODEs, what happens when you have variable coefficients?
E.g. the ODE xy′′+(2x−1)y′−2y=x2xy″+(2x−1)y′−2y=x2factorises to (xD−1)(D+2)[y]=x2(xD−1)(D+2)[y]=x2, so a particular solution ypyp should be ypyp = x2(xD−1)(D+2)x2(xD−1)(D+2). I can use the method outlined above for the D+2D+2 bit, but what about 1xD−11xD−1?
What does the inverse operator 1xD−11xD−1 mean?
If I were asked to find a particular solution, ypyp to (D−1)(D−2)[y]=x2ex(D−1)(D−2)[y]=x2ex, then I would use the formula g(x)D−a=eax∫e−axg(x)dxg(x)D−a=eax∫e−axg(x)dx, where g(x)=x2exg(x)=x2ex.
However, here's the difficulty I'm facing; all the ODE's that we have learned in class with operator methods have been dealing with constant coefficient ODEs, what happens when you have variable coefficients?
E.g. the ODE xy′′+(2x−1)y′−2y=x2xy″+(2x−1)y′−2y=x2factorises to (xD−1)(D+2)[y]=x2(xD−1)(D+2)[y]=x2, so a particular solution ypyp should be ypyp = x2(xD−1)(D+2)x2(xD−1)(D+2). I can use the method outlined above for the D+2D+2 bit, but what about 1xD−11xD−1?
What does the inverse operator 1xD−11xD−1 mean?
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