Asked by assma
solve d^3y/dx^3-3d^2y/dx^2+2dy/dx=exp(2x)
Answers
Answered by
Steve
y"' - 3y" + 2y' = e^(2x)
D(D-1)(D-2)(y) = e^2x
y = (1/4)(c1)e^x + (1/4)(c2)e^(2x) + (1/2)xe^(2x) + c3
D(D-1)(D-2)(y) = e^2x
y = (1/4)(c1)e^x + (1/4)(c2)e^(2x) + (1/2)xe^(2x) + c3
Answered by
assma
solve d^3y/dx^3+4d^2y/dx^2+4dy/dx=exp(-2x)
Answered by
Steve
that is just like the previous one. Just review your text section on linear DE's. There must be examples just like your problem.
Answered by
assma
how can find on the particular solution of the linear DE's
d^3y/dx^3+4d^2y/dx^2+4dy/dx=exp(-2x)
d^3y/dx^3+4d^2y/dx^2+4dy/dx=exp(-2x)
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