Based on the form of r(x) = − 6cos3x, our initial guess for the particular solution is yp(x) = Acos3x + Bsin3x (step 2). None of the terms in yp(x) solve the complementary

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We've got the following differential equation: y ″ ( x) + y ( x) = h ( x). I'm able to solve y ″ ( x) + y ( x) = 0 But I have no clue on how to solve the inhomogeneous equation. ordinary-differential

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Let the general solution of a second order homogeneous differential equation be \[{y_0}\left( x \right) = {C_1}{Y_1}\left( x \right) + {C_2}{Y_2}\left( x \right).\] Instead of the constants

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