Differential Equations: Homework Alvin Lin January 08 - May 08
Section.3 Exercise The direction field for provided x 0. dx = 4x y is shown. Verify that the straight lines y = ±x are solution curves, y = x dx = = 4x y = 4x x = y = x dx = = 4x y = Sketch the solution curve with initial condition y(0) =. 4x x = dx = 4x y y = 4x dx y = 4x dx y = x + c y(0) = c = 4 y = x + 4 5 4 4 5
Sketch the solution curve with initial condition y() =. dx = 4x y y = 4x dx y = 4x dx y = x + c y() = c = 7 y = x 7 4 4 4 4 What can you say about the behavior of the above solutions as x +? How about x? As x approaches positive/negative infinity, the solutions become more and more vertical. This makes sense since the slope approaches positive and negative infinity. Exercise 3 A model for the velocity v at time t of a certain object falling under the influence of gravity in a viscous medium is given by the equation dv dt = v 8 From the direction field, sketch the solutions with the initial conditions v(0) = 5, 8, 5. Why is the value v = 8 called the terminal velocity? 5 8 5 4 4 v = 8 is called the terminal velocity because no object can cross that line from either side. All objects will eventually accelerate or decelerate to v = 8 from any initial velocity. It represents the maximum velocity that an object can freefall at in the viscous medium. 3
Exercise 5 The logistic equation for the population (in thousands) of a certain species is given by: dp dt = 3p p If the initial population is 3000, what can you say about the limiting population lim t + p(t)? If p(0) = 0.8, what is lim t + p(t)? Can a population of 000 ever decline to 800? No, the population will stop decreasing at 500. lim p(t) = 500 t + lim p(t) = 500 t + Section.4 Exercise 5 Use Euler s method with step size h = 0. to approximate the solution to the initial value problem y = x y y() = 0 at the points x =.,.,.3,.4,.5. 4
x 0 = y 0 = x =. y = + 0.( ) = x =. y = + 0.(. ) =.0 x 3 =.3 y 3 =.0 + 0.(..0 ) =.08 x 4 =.4 y 4 =.08 + 0.(.3.08 ) =.05 x 5 =.5 y 5 =.05 + 0.(.4.05 ) =.08 Section. Exercise 3 Determine whether the given differential equation is separable. Exercise 7 Solve the equation: ds dt = t ln(st ) + 8t = t ln(s) + 8t = t (ln(s) + 4) ln(s) + 4 ds = t dt x x = y 3 y 3 = x dx y 3 = x dx y 4 = ln x + c 4 Exercise 9 Solve the equation: dx dt = t xe t+x xe t+x dx = t dt xe t e x dx = t dt xe x dx = t e dt t t xe x dx = e dt t xe x e x = te t e t + c 5
Exercise Solve the equation: dx = sec (y) + x sec (y) = + x dx cos (y) = x sin(y) cos(y) + + x dx = tan (x) + c Exercise 5 Solve the equation: (x + xy ) dx + e x y = 0 x( + y ) dx = e x y x dx = y e x + y x y dx = e x + y + c = ln y + e x Exercise 7 Solve the initial value problem dx = ( + y ) tan(x) with y(0) = 3: dx = ( + y ) tan(x) = tan(x) dx + y + y = tan(x) dx tan (y) = ln cos(x) + c tan ( 3) = ln cos(0) + c π 3 = c tan (y) = ln cos(x) + π 3 6
Exercise 9 Solve the initial value problem dx = y + cos(x) with y(π) = 0: dx = y + cos(x) = cos(x) dx y + y + = cos(x) dx y + = sin(x) + c = sin(π) + c c = y = (sin(x) + ) Exercise Solve the initial value problem θ = y sin θ dθ y + with y(π) = : θ dθ = y sin θ y + y + = θ sin θ dθ y y + = θ sin θ dθ y y + ln y = sin θ θ cos θ + c + ln = sin π π cos π + c + 0 = 0 + π + c c = π y + ln y = sin θ + θ cos θ + π 7
Exercise 3 Solve the initial value problem dt = t cos (y) with y(0) = π 4. dt = t cos (y) sec (y) = t dt sec (y) = t dt tan(y) = t + c tan( π 4 ) = 0 + c c = tan(y) = t + y = tan (t + ) Exercise 4 Solve the initial value problem dx = 8x3 e y with y() = 0. dx = 8x3 e y e y = 8x 3 dx e y = 8x 3 dx e y = x4 + c e 0 = (4 ) + c c = 3 e y = 4x 4 3 y = ln(4x4 3) 8
Exercise 6 Solve the initial value problem y dx + ( + x) = 0 with y(0) =. y dx + ( + x) = 0 ( + x) = y dx y = + x dx y = + x dx y = ln + x + c = ln + c = c y = ln + x ( ln + x y = If you have any questions, comments, or concerns, please contact me at alvin@omgimanerd.tech ) 9