Let Q1=y(1.1), Q2=y(1.2), Q3=y(1.3), where y=y(x) solves...
1) y'''+2y''-5y'- 6y=4x^2 where y(0)=1, y'(0)=2, y''(0)=3
2) y'''- 6y''+11y'-...
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Advance Math
Let Q1=y(1.1), Q2=y(1.2), Q3=y(1.3), where y=y(x) solves...
1) y'''+2y''-5y'- 6y=4x^2 where y(0)=1, y'(0)=2, y''(0)=3
2) y'''- 6y''+11y'- 6y=6e^(4x) where y(0)=4, y'(0)=10,y''(0)=30
3) y''- 6y'+9y=4e^(3x) ln(x) where y(1)=, y'(1)=2
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1 Find the complementary solution by solving d3 yx dx3 2 d2 yx dx2 5 dyx dx 6 yx 0 Assume a solution will be proportional to e x for some constant Substitute yx e x into the differential equation d3 dx3e x 2 d2 dx2e x 5 d dxe x 6 e x 0 Substitute d3 dx3e x 3 e x d2 dx2e x 2 e x and d dxe x e x 3 e x 2 2 e x 5 e x 6 e x 0 Factor out e x 3 2 2 5 6 e x 0 Since e x 0 for any finite the zeros must come from the polynomial 3 2 2 5 6 0 Factor 2 1 3 0 Solve for 3 or 1 or 2 The root 3 gives y1x c1 e3 x as a solution where c1 is an arbitrary constant The root 1 gives y2x c2 ex as a solution where c2 is an arbitrary constant The root 2 gives y3x c3 e2 x as a solution where c3 is an arbitrary constant The general solution is the sum of the above solutions yx y1x y2x y3x c1 e3 x c2 ex c3 e2 x Determine the particular solution to d3 yx dx3 2 d2 yx dx2 5 dyx dx 6 yx 4 x2 by the method of undetermined coefficients The particular solution to d3 yx dx3 2 d2 yx dx2 5 dyx dx 6 yx 4 x2 is of the form ypx a1 a2 x a3 x2 Solve for the unknown constants a1 a2 and a3 Compute dypx dx dypx dx d dxa1 a2 x a3 x2 a2 2 a3 x Compute d2 ypx dx2 d2 ypx dx2 d2 dx2a1 a2 x a3 x2 2 a3 Compute d3 ypx dx3 d3 ypx dx3 d3 dx3a1 a2 x a3 x2 0
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