# 2_3a - Solutions 2.3-Page 131 Problem 1 Find the general...

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Solutions 2.3-Page 131 Problem 1 Find the general solutions of the differential equations. 0 4 = y y The first step is to find the roots of the characteristic equation. 2 , 2 0 ) 2 )( 2 ( 0 4 2 = = + = r r r r Since the roots are real and distinct, the general solution is x x e c e c x y 2 2 2 1 ) ( + =

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Problem 9 Find the general solutions of the differential equations. 0 25 8 = + + y y y The characteristic equation is . The roots can be found using the quadratic equation formula or a calculator. 0 25 8 2 = + + r r The roots are . i r 3 4 ± = The general solution is based on Theorem 3 on page 128. ) 3 sin 3 cos ( ) ( 2 1 4 x c x c e x y x + =
Problem 21 Solve the initial value problems. 4 3 0; (0) 7, (0) 11 yyy y y ′′ −+= = = The characteristic equation is . The roots can be found using the quadratic equation formula or a calculator. 0 3 4 2 = + r r The roots are . 1 , 3 = r The general solution is . The initial conditions are used to find the constants.

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2_3a - Solutions 2.3-Page 131 Problem 1 Find the general...

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