# Differential Equations and Linear Algebra Stephen W. Goode by Stephen W. Goode and Scott A. Annin By Stephen W. Goode and Scott A. Annin

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Extra resources for Differential Equations and Linear Algebra Stephen W. Goode and Scott A. Annin SOLN MANUAL

Example text

Sin−1 ( v4 ) = ln |x| + c =⇒ sin−1 ( 4x x 13. We first rewrite the given differential equation in the equivalent form y = (9x2 + y 2 ) + y . Factoring x |x| 9 + ( xy )2 + y . Since we are told to solve the differential x equation on the interval x > 0 we have |x| = x, so that y = 9 + ( xy )2 + xy , which we recognize as being homogeneous. We therefore√let y = xV , so that y = xV√ + V . Substitution into the preceding differential equation yields xV + V = 9 + V 2 + V , that is xV = 9 + V 2 . Separating the variables in this equation √ 1 1 we obtain √ dV = dx.

Tx − 1 x2 + y 2 = f (x, y). Thus f is homogeneous of degree x−y 56 5(ty) + 9 3(tx) + 5(ty) 3x + 5y tx − 3 + = = = f (x, y). Thus f is homogeneous of degree ty 3(ty) 3(ty) 3y y 3 + 5x 3x + 5y 3 + 5v zero. f (x, y) = = = = F (v). 3y 3 xy 3v 6. f (tx, ty) = (tx)2 + (ty)2 x2 + y 2 = = f (x, y). Thus f is homogeneous of degree zero. f (x, y) = tx x y √ |x| 1 + ( x )2 y 2 x2 + y 2 =− 1+ = − 1 + v 2 = F (v). = x x x 7. f (tx, ty) = (tx)2 + 4(ty)2 − (tx) + (ty) x2 + 4y 2 − x + y = f (x, y). Thus f is homogeneous of = x + 3y (tx) + 3(ty) √ 1 + 4( xy )2 − 1 + xy 1 + 4v 2 − 1 + v x2 + 4y 2 − x + y = F (v).

Dx (b) If n = 1, 2, then = [v01−n + (n − 1)kt]1/(1−n) , where x(t) denotes the distanced traveled by the object. dt 1 Consequently, x(t) = − [v 1−n + (n − 1)kt](2−n)/(1−n) + c. Imposing the initial condition x(0) = 0 k(2 − n) 0 1 1 1 v 2−n , so that x(t) = − [v 1−n + n(n − 1)kt](2−n)/(1−n) + v 2−n . For yields c = k(2 − n) 0 k(2 − n) o k(2 − n) 0 2−n 1 < 0, so that limt→∞ x(t) = . Hence the maximum distance that the 1 < n < 2, we have 1−n k(2 − n) 1 object can travel in a finite time is less than .

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