Exam 1: Introduction to Differential Equations

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The population of a town increases at a rate proportional to its population. Its initial population is 1000. The correct initial value problem for the population, P(t)P ( t ) , as a function of time, t, is

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The values of c for which y=cy = c is a constant solution of y′=y2+5y−6y ^ { \prime } = y ^ { 2 } + 5 y - 6 are

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The differential equation y′+3y=sin⁡xy ^ { \prime } + 3 y = \sin x is

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The solution of the initial value problem y′=2y+x,y(1)=1/4y ^ { \prime } = 2 y + x , y ( 1 ) = 1 / 4 is y=−x/2−1/4+ce2xy = - x / 2 - 1 / 4 + c e ^ { 2 x } , where c=c =

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A large mixing tank initially contains 100 gallons of water in which 30 pounds of salt have been dissolved. Another brine solution is pumped into the tank at the rate of 4 gallons per minute, and the resulting mixture is pumped out at the same rate. The concentration of the incoming brine solution is 2 pounds of salt per gallon. If A(t)A ( t ) represents the amount of salt in the tank at time t, the correct differential equation for A is

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The values of m for which y=emxy = e ^ { m x } is a solution of y′′−6y′−7y=0y ^ { \prime \prime } - 6 y ^ { \prime } - 7 y = 0 are

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The values of m for which y=xmy = x ^ { m } is a solution of x2y′′−7xy′+12y=0x ^ { 2 } y ^ { \prime \prime } - 7 x y ^ { \prime } + 12 y = 0 are

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In the previous problem, over a long period of time, the total amount of salt in the tank will approach

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The solution of the initial value problem y′=2y+x,y(−1)=1/2y ^ { \prime } = 2 y + x , y ( - 1 ) = 1 / 2 is y=−x/2−1/4+ce2xy = - x / 2 - 1 / 4 + c e ^ { 2 x } , where c=c =

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In the falling body problem, the units of acceleration might be

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The values of m for which y=emxy = e ^ { m x } is a solution of y′′−4y′−5y=0y ^ { \prime \prime } - 4 y ^ { \prime } - 5 y = 0 are

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The values of c for which y=cy = c is a constant solution of y′=y2+3y−4y ^ { \prime } = y ^ { 2 } + 3 y - 4 are

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In the previous problem, after a long period of time, the temperature of the coffee approaches

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The solution of the initial value problem y′=5y,y(1)=3y ^ { \prime } = 5 y , y ( 1 ) = 3 is y=ce5xy = c e ^ { 5 x } , where c=c =

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The differential equation y′′′+2y′′+3xy′−4exy=sin⁡xy ^ { \prime \prime \prime } + 2 y ^ { \prime \prime } + 3 x y ^ { \prime } - 4 e ^ { x } y = \sin x is

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The differential equation y′′+2y′+3y=sin⁡yy ^ { \prime \prime } + 2 y ^ { \prime } + 3 y = \sin y is

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The initial value problem y′=y2−9,y(x0)=y0y ^ { \prime } = \sqrt { y ^ { 2 } - 9 } , y \left( x _ { 0 } \right) = y _ { 0 } has a unique solution guaranteed by Theorem 1.1 if

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The differential equation y′′′+2y′′+3xy′−4exy=sin⁡xy ^ { \prime \prime \prime } + 2 y ^ { \prime \prime } + 3 x y ^ { \prime } - 4 e ^ { x } y = \sin x is

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The differential equation y′′+2yy′+3y=0y ^ { \prime \prime } + 2 y y ^ { \prime } + 3 y = 0 is

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The values of m for which y=emxy = e ^ { m x } is a solution of y′′−5y′+6y=0y ^ { \prime \prime } - 5 y ^ { \prime } + 6 y = 0 are

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