{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MA266 prac_ex1

# MA266 prac_ex1 - 1 A solution of dy 2y = with y(1 = 8 is dt...

This preview shows pages 1–3. Sign up to view the full content.

1. A solution of dy dt = 2 y t + 1 with y (1) = 8 is A. y = ( t + 1) 2 + 4 B. y = 32( t + 1) - 2 C. y = 2 ( t + 1) 2 D. y = 4 p 2( t + 1) E. y = p ( t + 1) 2 + 60 2. An implicit solution of y 0 = 2 x y + x 2 y is A. y 2 = 2 ln(1 + x 2 ) + C B. y 2 = C ln(1 + x 2 ) C. 1 2 y 2 = ln x 2 + C D. y 2 = ln(1 + x 2 ) + C E. 1 2 y 2 = ln | 1 + x | + C 3. The substitution v = y/x transforms the equation dy dx = sin( y/x ) into A. v 0 = sin( v ) B. v 0 = x sin( v ) C. v 0 + v = sin( v ) D. xv 0 + v = sin( v ) E. v 0 + xv = sin( v ) 4. The solution in implicit form of dy dx = x 2 + 3 y 2 2 xy is: A. x 2 + y 2 = x 3 + C B. x 2 + y 2 = Cx 3 C. x 2 + x 3 = y 2 + C D. Cx 2 = x 3 + y 2 E. x 2 + y 3 + xy 2 = C 5. Which of the following best describes the stability of equilibrium solutions for the au- tonomous differential equation y 0 = y (4 - y 2 )? A. y = 0 unstable; y = 2 and y = - 2 both stable 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
B. y = 0 unstable; y = 2 stable C. y = 0 and y = 2 both stable D. y = 0 stable; y = 2 unstable; y = - 2 stable E. y = 0 stable; y = - 2 and y = 2 both unstable 6. Solve the initial value problem y 0 - 2 y = e - 2 t with y (0) = a . For what value(s) of a is the solution bounded (i.e., not tending to infinity as t + ) on the interval t > 0?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern