Solution Manual for Advanced Engineering.pdf

Chapter 1 First-Order Differential Equations 1.1 Terminology and Separable Equations 1. The differential equation is separable because it can be written 3y 2 dy = 4x, dx or, in differential form, Integrate to obtain 3y 2 dy = 4xdx. y 3 = 2x 2 + k. This implicitly defines a general solution, which can be written explicitly as y = (2x 2 + k) 1/3 , with k an arbitrary constant. 2. Write the differential equation as dy x dx = −y, which separates as 1 1 if x /= 0 and y y dy = − x dx 0. Integrate to get ln |y| = − ln |x| + k. Then ln |xy| = k, so xy = c 1 . − ln. y − 1. = k. 2 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS with c constant (c = e k ). y = 0 is a singular solution, satisfying the original differential equation. 3. If cos(y) /= 0, the differential equation is y sin(x + y) = dx cos(y) sin(x) cos(y) + cos(x) sin(y) = cos(y) = sin(x) + cos(x) tan(y). There is no way to separate the variables in this equation, so the differential equation is not separable. 4. Write the differential equation as e x e y dy = 3x, dx which separates in differential form as e y dy = 3xe−x dx. Integrate to get e y = −3e −x (x + 1) + c, with c constant. This implicitly defines a general solution. 5. The differential equation can be written x dy = y 2 y, dx or 1 y(y − 1) dy = 1 dx, x and is therefore separable. Separating the variables assumes that y /= 0 and y 1. We can further write 1 − 1 dy = 1 dx. Integrate to obtain y − 1 y x ln |y − 1| − ln |y| = ln |x| + k. Using properties of the logarithm, this is . xy . . / / − 2 y 2y 2 + 1 x x + 1 1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 3 Then y − 1 = c, xy with c = e k constant. Solve this for y to obtain the general solution 1 y = . 1 − cx y = 0 and y = 1 are singular solutions because these satisfy the differential equation, but were excluded in the algebra of separating the variables. 6. The differential equation is not separable. 7. The equation is separable because it can be written in differential form as sin(y) cos(y) dy = 1 dx. x This assumes that x /= 0 and cos(y) /= 0. Integrate this equation to obtain — ln | cos(y)| = ln |x| + k. This implicitly defines a general solution. From this we can also write sec(y) = cx with c constant. The algebra of separating the variables required that cos(y) 0. Now cos(y) = 0 if y = (2n+1)π/2, with n any integer. Now y = (2n+1)π/2 also satisfies the original differential equation, so these are singular solutions. 8. The differential equation itself requires that y = 0 and x = 1. Write the equation as x dy = y dx and separate the variables to get 1 2y 2 + 1 x 1 y(2y 2 + 1) dy = x(x + 1) dx. Use a partial fractions decomposition to write this as 1 − 2y dy = 1 − 1 dx. Integrate to obtain ln |y| − 1 ln(1 + 2y 2 ) = ln |x| − ln |x + 1| + c . − 2 1 + 2y 2 x + 1 1 + 2y 2 x + 1 4 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS with c constant. This implicitly defines a general solution. We can go a step further by writing this equation as ln √ y ! = ln x + c and take the exponential of both sides to get √ y = k x , which also defines a general solution. 9. The differential equation is dy = e x y + sin(y), dx and this is not separable. It is not possible to separate all terms involving x on one side of the equation and all terms involving y on the other. 10. Substitute sin(x − y) = sin(x) cos(y) − cos(x)sin(y), cos(x + y) = cos(x) cos(y) − sin(x) sin(y), and cos(2x) = cos2 (x) − sin2 (x) into the differential equation to get the separated differential form (cos(y) − sin(y)) dy = (cos(x) − sin(x)) dx. Integrate to obtain the implicitly defined general solution cos(y) + sin(y) = cos(x) + sin(x) + c. 11. If y /= −1 and x /= 0, we obtain the separated equation y 2 y + 1 dy = 1 dx. x To make the integration easier, write this as y − 1 + 1 dy = 1 dx. 1 + y x Integrate to obtain 1 y 2 − y + ln |1 + y| = ln |x| + c.