By Kreyszig E.

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**Extra resources for Solutions manual for Advanced engineering mathematics 9ed.**

**Example text**

4m(m Ϫ 1) ϩ 4m Ϫ 1 ϭ 4(m Ϫ _12)(m ϩ _12) ϭ 0, c1͙xෆ ϩ c2/͙xෆ 4. (c1 ϩ c2 ln x)/x 6. 52] ϭ 0. 5i. 5 ln ͉x͉)]. 8. 4(m(m Ϫ 1) ϩ _14) ϭ 4(m Ϫ _12)2 ϭ 0 has the double root m ϭ _12; hence a general solution is y ϭ (c1 ϩ c2 ln ͉x͉)͙xෆ. 10. 12] ϭ 0. 1 ln ͉x͉)]. 12. The auxiliary equation is m(m Ϫ 1) ϩ 3m ϩ 1 ϭ m2 ϩ 2m ϩ 1 ϭ (m ϩ 1)2 ϭ 0. It has the double root Ϫ1. Hence a general solution is y ϭ (c1 ϩ c2 ln ͉x͉)/x. The first initial condition gives y(1) ϭ c1 ϭ 4. The derivative of y is yЈ ϭ c2 /x 2 ϩ (c1 ϩ c2 ln ͉x͉)/(Ϫx 2).

We differentiate this with respect to x and equate the result to the coefficient of dx in the exact ODE. This gives ux ϭ 2xy ϩ y 2 ϩ lЈ ϭ 2xy ϩ y 2 ϩ e x(x ϩ 1); hence lЈ ϭ e x(x ϩ 1) and by integration, l ϭ xe x. The general implicit solution is u(x, y) ϭ x 2y ϩ xy 2 ϩ xe x ϭ c. From the initial condition, u(1, 1) ϭ 1 ϩ 1 ϩ e ϭ 2 ϩ e. The particular solution of the initial value problem is u(x, y) ϭ x 2y ϩ xy 2 ϩ xe x ϭ 2 ϩ e. Instructor’s Manual 27 34. In problems of this sort we need two conditions, because we must determine the arbitrary constant c in the general solution and the constant k in the exponent.

Y1 ϭ e 2x cos x, y2 ϭ e 2x sin x, W ϭ e 4x, so that (2) gives ͵ sin x ͵ e yp ϭ Ϫe 2x cos x e2x sin x e 2x csc x e؊4x dx ϩ e 2x 2x cos x e 2x csc x e؊4x dx ϭ Ϫe 2x (cos x) x ϩ e 2x (sin x) ln ͉sin x͉. 56 Instructor’s Manual 18. This is an Euler–Cauchy equation, with the right-side 0 replaced with 36x 5. 5)2 ϭ 0. 5(c1 ϩ c2 ln ͉x͉). We try yp ϭ Cx 5. y pЈ ϭ 5Cx 4, Then y pЉ ϭ 20Cx 3. Substitution gives (4 ⅐ 20 Ϫ 24 ⅐ 5 ϩ 49)Cx 5 ϭ 9Cx 5 ϭ 36x 4; hence C ϭ 4. 5(c1 ϩ c2 ln ͉x͉) ϩ 4x 5 of the nonhomogeneous ODE.