Difference between revisions of "Tutorial Week 1"

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## $x y'+3 y=x, \quad y(1)=0$
 
## $x y'+3 y=x, \quad y(1)=0$
 
## $x y'+3 y=x^{-1} e^x$
 
## $x y'+3 y=x^{-1} e^x$
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## $y'=(1 + y^2)(3x^2-1)$  <noinclude> -- Remove this for next year (separable equations aren't covered until week 2)</noinclude>
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## $y'=(1 + y^2)(3x^2-1)$  <includeonly> -- Remove this for next year (separable equations aren't covered until week 2)</includeonly>
  
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=== Solutions ===
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[[Tutorial Week 1 Solutions]]
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## 3rd order, linear, ODE '''(3 pts)'''
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## 2nd order in space, 1st order in time, linear, PDE '''(4 pts)'''
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## 2nd order, nonlinear, ODE '''(3 pts)'''
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## $A>3$ '''(1 pt)'''
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## $A<3$ '''(1 pts)'''
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## $A = 3$ '''(2 pt)'''
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## [https://www.desmos.com/calculator/9oztnvu137 Graph] '''(3 pts)'''<!--[http://graphsketch.com/?eqn1_color=1&eqn1_eqn=2-x%2B(4-2)*exp(5*x)&eqn2_color=2&eqn2_eqn=2-x%2B(3-2)*exp(5*x)&eqn3_color=3&eqn3_eqn=2-x%2B(2-2)*exp(5*x)&eqn4_color=4&eqn4_eqn=2-x%2B(1-2)*exp(5*x)&eqn5_color=5&eqn5_eqn=2-x%2B(0-2)*exp(5*x)&eqn6_color=6&eqn6_eqn=&x_min=-1&x_max=1&y_min=-10.5&y_max=10.5&x_tick=1&y_tick=1&x_label_freq=5&y_label_freq=5&do_grid=0&do_grid=1&bold_labeled_lines=0&bold_labeled_lines=1&line_width=4&image_w=850&image_h=525 Graphs]-->
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## $y = \frac{x}{4} - \frac{1}{4x^3}$. Either multiply through by $x^2$, or divide through by $x$ and find the integrating factor ($x^3$) '''(1 pt)''', take antiderivatives  on both sides '''(2 pts)''', solve for $y$ '''(1 pt)''', use IC to determine the value of the arbitrary constant '''(1 pt)'''. Sketch - must show vertical and slant asymptote '''(2 pts)'''.
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## $y = \frac{e^x(x - 1) + C}{x^3}$. Same as above (no points for repeated steps) except that the antiderivative of $xe^x$ requires integration by parts '''(2 pts)'''. Must include the arbitrary constant '''(1 pt)''' and solve for $y$ '''(1 pt)'''.
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## $y = \tan(x^3 - x + c)$. The equation is separable. Separate variables '''(1 pt)''', take antiderivatives on both sides '''(2 pts)''', solve for $y$ '''(1 pt)'''. For the solution checking process, $y'=\sec^2(x^3 - x + c)(3x^2 - 1)$ and $(1 + y^2) (3x^2 - 1) = (1 + \tan^2(x^3 - x + c))(3 x^2 - 1)$ which are equal using trig identities '''(2 pts)'''.
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Latest revision as of 22:57, 30 December 2020

Worksheet Questions

  1. For each of the following differential equations, state whether it is an ODE or a PDE, state its order (if it’s a PDE, give the order in each independent variable) and whether it is linear or nonlinear.
    1. $t^3y'''+y'=\sin(t)$
    2. $\frac{\partial^2 y}{\partial x^2} + x^2 \frac{\partial y}{\partial t} = x^3$
    3. $e^x y+y \frac{d^2y}{dx^2} = x\cos(x)$
  2. Suppose that $y(t)=2+1/(1+t)+Ce^{3t}$ is the general solution to some differential equation. Consider the initial condition $y(0)=A$.
    1. For what values of $A$ does $\lim_{t\to+\infty}y(t) = \infty$?
    2. For what values of $A$ does $\lim_{t\to+\infty}y(t) = -\infty$?
    3. What, if any, are the values of $A$ for which this limit is finite?
    4. Sketch 3 members of this family of functions choosing values of A so that the behaviour of all types of solutions is demonstrated.
  3. Find solutions to the following ordinary differential equations. Recall that you can always check that you got the correct solution by plugging it back in to the given differential equation. Sketch the solution to the initial-value problem in part a. Show solution-checking process for the solution to the equation in part c.
    1. $x y'+3 y=x, \quad y(1)=0$
    2. $x y'+3 y=x^{-1} e^x$
    3. $y'=(1 + y^2)(3x^2-1)$

Tutorial Week 1 Solutions

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