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The general solution of Y "+ y = x + cosx
Y "+ y = x + cosx first calculate Y" + y = 0, get y = asinx + bcosx, a, B are constants, then calculate the special solution, the special solution can be divided into two parts (using the linear relationship), one part is the special solution whose calculation result is equal to x, the other part is the special solution whose calculation result is equal to cosx 1, the calculation result is equal to the special solution of X
Find the general solution of the differential equation y "+ y = cosx
The characteristic equation of the original equation corresponding to the homogeneous equation y '' + y = 0 is: R2 + 1 = 0, and its characteristic roots are: R1 = I, R2 = - I, so the general solution of the homogeneous equation is: y = c1cosx + c2sinx. Let a special solution of the non-homogeneous equation y '' + y = cosx be: y2 = excosx + dxsinx. Substituting it into the equation, e = 0, d = 12. So y2 = 12xsinx. So the general solution of the original equation is y = c1cosx + c2sinx + 12xsinx
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