Question

The particular integral a tre diferential equation( D''+ D + 1) = sin2x is​

Answer 1

Step-by-step explanation:

1/f(D)]eax = [1/f(a)]eax If f(a) = 0 then [1/f(D)]eax = x[1/f'(a)]eax If f'(a) = 0 then [1/f(D)]eax = x2[1/f''(a)]eax

[1/f(D)]xn = [f(D)]-1xn expand [f(D)]-1 and then operate.

[1/f(D2)]sin ax = [1/f(-a2)]sin ax. and [1/f(D2)]cos ax = [1/f(-a2)]cos ax. ...

[1/f(D)]eax φ(x) = eax [1/f(D+a)]φ(x)

[1/(D+a)]φ(x) = e-ax∫eaxφ(x) dx.

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