Identify significance of terms The following differential equation models the behaviour of a damped-driven oscillator, where \(x\) is position and \(t\) is time.
\[ \frac{\mathrm{d}^2 x}{\mathrm{d} t^2} + k \frac{\mathrm{d} x}{\mathrm{d} t} +\omega^2 x = p(t). \]
What is the physical significance of each term?

]]> 1.0000000 0.3333333 0 true Your answer is correct.

]]> Your answer is partially correct.

]]> Your answer is incorrect.

]]> \(\frac{\mathrm{d}^2 x}{\mathrm{d} t^2} \)

]]> Accelleration \(k \frac{\mathrm{d} x}{\mathrm{d} t}\)

]]> Friction/damping \(\omega^2 x\)

]]> Force \(p(t)\)

]]> External forcing Categorize Describe the following differential equation accurately
\[ 7\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + y^2 \frac{\mathrm{d}y}{\mathrm{d}t}-3\sin(t)=0.\]

]]> 1.0000000 0.3333333 0 false true abc Your answer is correct.

]]> Your answer is partially correct.

]]> Your answer is incorrect.

]]> Linear

]]> Second order

]]> First order

]]> Homogeneous

]]> Inhomogeneous

]]> Nonlinear

]]> Which method to solve Which method would you use to solve the following differential equation?

\[\displaystyle \frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{e^{-x^2}}{y}-2x,\quad y(0)=1. \]

]]> 1.0000000 0.3333333 0 true true abc Your answer is correct.

]]> Your answer is partially correct.

]]> Your answer is incorrect.

]]> Linear ODE, so use the substitution \(x(t)=e^{\lambda t}\).

]]> Laplace Transforms.

]]> Integrating factor.

]]> It is exact, so reformulate and integrate directly.

]]> None of these methods - it has no solution in terms of elementary functions.

]]> 1st-Order-ODE-1-sep Solve the following differential equation: \[ {@v@}\frac{\mathrm{d}y}{\mathrm{d}{@v@}} + {@n2@} {@v@}+{@n1@} = 0. \] Use the letter "c" (small case) to denote the arbitrary constant in the general solution.
\(y({@v@})\) =[[input:ans1]] [[validation:ans1]]

]]> For the equation \[ {@v@}\frac{\mathrm{d}y}{\mathrm{d}{@v@}} + {@n2@}{@v@}+{@n1@} = 0 \] rearranging gives \[\mathrm{d}y = \frac{(-{@n2@}{@v@}-{@n1@})}{@v@}\mathrm{d}{@v@}\] Therefore \[\int \mathrm{d}y = \int\frac{(-{@n2@}{@v@}-{@n1@})}{@v@}d{@v@} =\int(-{@n2@}-\frac{{@n1@}}{{@v@}})\mathrm{d}{@v@}\] Hence \[y = {@correctanswer@}\]

]]> 2.0000000 0.0000000 0 n1:rand(8)+2; n2:rand(6)+2; v:x; fans:-n2-n1/v; correctanswer:int(fans,v)+c; wronganswer:-n1*ln(v)-n2*v+c; [[feedback:Result]]

]]> \({@v@}\frac{\mathrm{d}y}{\mathrm{d}{@v@}} + {@n2@}{@v@}+{@n1@} = 0\) gives {@correctanswer@} 1 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> none 1 i cos-1 [ ans1 algebraic correctanswer 20 1 0 0 diff 1 1 1 1 1 Result 2.0000000 1 0 AlgEquiv ans1 correctanswer 0 = 1.0000000 -1 Result-0-T = 0.0000000 1 Result-0-F 1 AlgEquiv diff(ans1,x) fans 0 = 0.7500000 2 Result-1-T Your answer satisfies the differential equation, but is not in the correct form.

]]> = 0.0000000 -1 Result-1-F 2 AlgEquiv member(c,listofvars(ans1)) true 0 + 0.2500000 -1 Result-3-T + 0.0000000 -1 Result-3-F Your answer should have the general constant \(c\), but does not.

]]> 1645139326 481337292 1509361853 346176631 915142823 963104251 793273982 1455145271 1 ans1 correctanswer Result 1.0000000 0.0000000 Result-0-T 2 ans1 ev(correctanswer-c,simp) Result 0.7500000 0.0000000 Result-3-F 3 ans1 wronganswer Result 1.0000000 0.0000000 Result-3-T 4 ans1 correctanswer-c+Q Result 0.7500000 0.0000000 Result-3-F 5 ans1 0 Result 0.0000000 0.0000000 Result-1-F 1st-Order-ODE-2-sep Solve the following differential equation: \[ {@sqrt(n1-x^2)@}\frac{\mathrm{d}y}{\mathrm{d}x}-x(y+{@n2@})=0\] Use the letter \(c\) to denote the arbitrary constant in the general solution.
\(y(x)\) = [[input:ans1]] [[validation:ans1]]

]]> For the equation \[ {@sqrt(n1-x^2)@}\frac{\mathrm{d}y}{\mathrm{d}x} -x(y+ {@n2@}) = 0 \] rearranging gives \[\frac{\mathrm{d}y}{y+ {@n2@}} = \frac{x\mathrm{d}x}{{@sqrt(n1-x^2)@}}\] To integrate, let \(u = {@n1@}-x^2\) giving \(du = -2dx\) \[\int \frac{\mathrm{d}y}{y+ {@n2@}} = \int\frac{-\mathrm{d}u}{2{@sqrt(u)@}}\]

Hence \[\ln(|y+{@n2@}|) = {@-u^(1/2)@}+c\] and \[y + {@n2@} = {@e^(-u^(1/2)+c)@} = {@e^(c-sqrt(n1-x^2))@}=c{@e^(-(n1-x^2)^(1/2))@}\]
Hence \[y = c{@e^(-(n1-x^2)^(1/2))@}-{@n2@}\]

]]> 3.0000000 0.0000000 0 n1:rand(8)+2; n2:rand(6)+2; correctanswer:c*e^(-(n1-x^2)^(1/2))-n2; correctanswer2:e^(c-(n1-x^2)^(1/2))-n2; wronganswer1:Q*e^(-(n1-x^2)^(1/2))-n2; wronganswer2:exp(c)*exp(-(n1-x^2)^(1/2))-n2; wronganswer3:e^(-(n1-x^2)^(1/2)+c)-n2; check2:x/sqrt(n1-x^2); [[feedback:Result]]

]]> \({@sqrt(n1-x^2)@}\frac{\mathrm{d}y}{\mathrm{d}x} -x(y+ {@n2@}) = 0\) gives \({@correctanswer@}\) 1 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> none 1 i cos-1 [ ans1 algebraic correctanswer 20 1 0 0 1 1 1 1 1 Result 3.0000000 1 studentderivcheck:ev(diff(ans1,x)/(ans1+n2),simp); 0 AlgEquiv ans1 correctanswer 0 = 1.0000000 -1 Result-0-T = 0.0000000 1 Result-0-F 1 AlgEquiv ans1 correctanswer2 0 = 1.0000000 0.0000000 -1 Result-1-T = 0.0000000 2 Result-1-F 2 AlgEquiv studentderivcheck check2 0 = 0.7500000 -1 Result-3-T Your answer satisfies the differential equation, but is not in the desired simplest format.

]]> - 0.0000000 -1 Result-3-F 1355936822 145020125 1715062959 840751671 462422398 1959186500 1564539276 1919149652 662492552 901197713 832113321 2111734750 455246304 1 ans1 correctanswer Result 1.0000000 0.0000000 Result-0-T 2 ans1 correctanswer+n2 Result 0.0000000 0.0000000 Result-3-F 3 ans1 wronganswer1 Result 0.7500000 0.0000000 Result-3-T 4 ans1 wronganswer2 Result 1.0000000 0.0000000 Result-1-T 5 ans1 wronganswer3 Result 1.0000000 0.0000000 Result-1-T 1st-Order-ODE-3-exact Solve the following differential equation \[ {@ode@}=0.\] Enter your answer as an explicit function \(y(x)\).

\(y(x)=\) [[input:ans1]] [[validation:ans1]]

]]> We are trying to solve \({@ode@}=0\). First we check if this equation is exact. It is already written in the form \[ p(x,y)\cdot \dot{y}(x) + q(x,y) =0.\] Here \( p(x,y) = {@p@}\) and \(q(x,y) = {@q@}.\) Assume that \(h(x,y)=c\) gives an implicit function, which satisfies this equation. Then \[ \frac{\mathrm{d}h}{\mathrm{d}x}=\frac{\partial h}{\partial y}\cdot \frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\partial h}{\partial x}=0\] and so \[ \frac{\partial h}{\partial y} = p(x,y), \quad \frac{\partial h}{\partial x}=q(x,y).\] Differentiating once further we have \[ \frac{\partial p}{\partial x} = \frac{\partial^2 h}{\partial x\partial y}=\frac{\partial q}{\partial y}.\] Note that this condition on \(p\) and \(q\) is necessary and sufficient for the ODE to be exact. \[ \frac{\partial p}{\partial x} = {@diff(p,x)@}\] and \[\frac{\partial q}{\partial y} = {@diff(q,y)@}.\] Since these are equal the ODE is exact. Now we try to find the function \(h(x,y)\). \[ h_1 = \int q(x,y)\mathrm{d}x + c_1(y) = \int {@q@}\mathrm{d}x + c_1(y) = {@int(q,x)@}+ c_1(y),\] \[ h_2 = \int p(x,y)\mathrm{d}y + c_2(x) = \int {@p@}\mathrm{d}y + c_2(x) = {@int(p,y)@}+c_2(x).\] Notice here that \(c_1\) and \(c_2\) are arbitrary functions of integration. Equating these two expressions to find these functions we ultimately get \[ h(x,y) = {@h@}\] The solution to the ODE is therefore \[ {@h@}=c\] where \(c\) is an arbitrary constant of integration. In this case we can solve for \(y\) to get an explicit solution, giving \[ y(x) = {@ta@}.\]

]]> 3.0000000 0.0000000 0 /* NOTE: make sure any randoms don't break the solve in the second line... */ h:x*y+(rand(5)+2)*x^2 ta:rhs(first(solve(h+c=0,y))) p:diff(h,y) q:diff(h,x) ode:p*'diff(y,x)+q [[feedback:Result]]

]]> \(h(x,y)=\) {@h@} giving {@ode@} and solution {@ta@}. 1 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> none 1 i cos-1 [ ans1 algebraic ta 20 1 0 0 1 1 1 1 1 Result 3.0000000 1 sa1 : subst(y=ans1,ode); sa2 : ev(sa1,nouns); l:listofvars(ans1) 0 AlgEquiv member(y,listofvars(ans1)) true 0 = 0.0000000 0.0000000 -1 Result-1-T Your answer should be an explicit function, but yours appears to depend on \(y\)!

]]> = 0.0000000 1 Result-1-F 1 AlgEquiv sa2 0 0 = 1.0000000 0.0000000 2 Result-2-T = 0.0000000 -1 Result-2-F Your answer does not satisfy the differential equation:

\[ \left({@p@}\right)\frac{\mathrm{d}y}{\mathrm{d}x}+\left({@q@}\right)=\left({@p@}\right){\times}\left({@diff(ans1,x)@}\right) +{@q@} = {@sa2@} = {@expand(sa2)@}.\]

]]> 2 AlgEquiv length(l) 2 0 + 0.0000000 -1 Result-3-T = 0.5000000 -1 Result-3-F Your answer should have an arbitrary constant, but does not.

]]> 1151396001 1523168427 978880811 1993545210 1533657032 1 ans1 ta Result 1.0000000 0.0000000 Result-3-T 2 ans1 rhs(first(solve(h=0,y))) Result 0.5000000 0.0000000 Result-3-F 3 ans1 h+c Result 0.0000000 0.0000000 Result-1-T 2nd-Order-ODE-1-real-distinct Solve \[{@n1@} \frac{d^2y}{dt^2} - {@bc@} \frac{dy}{dt} = {@cc@}y .\]
\(y(t)\) = [[input:ans1]] [[validation:ans1]]

]]> To solve the differential equation \[{@n1@} \frac{d^2y}{dt^2} - {@bc@} \frac{dy}{dt} = {@cc@} y. \]
First write this in standard form \[{@n1@} \frac{d^2y}{dt^2} - {@bc@} \frac{dy}{dt} - {@cc@} y = 0\]
Let \(y = e^{mt}\) and hence \(\frac{dy}{dt}=me^{mt}\) and \(\frac{d^2y}{dt^2} = m^2e^{mt}\).
Substituting for \(y\) gives \[{@n1@}m^2e^{mt} - {@bc@}me^{mt} - {@cc@}e^{mt} = 0\] and dividing throughout by \(e^{mt}\) and {@n1@} gives \[m^2 - {@bc/n1@}m - {@cc/n1@} = 0\] which can be solved to give {@m@}. The solution is then of the form \(y = A{@e^(m1*t)@}+B{@e^(m2*t)@}\).

]]> 2.0000000 0.0000000 0 n1:rand(8)+2; n2:rand(6)+6; n3:rand(5)+1; bc:n1*(n2-n3); cc:n1*n2*n3; m:solve(n1*m^2 - bc*m - cc = 0,m); m1:rhs(m[1]); m2:rhs(m[2]); correctanswer:A*e^(m1*t)+B*e^(m2*t); altanswer:B*e^(m1*t)+A*e^(m2*t); wronganswer2:A*e^(m1*t); q:n1*'diff(y(t),t,2)-bc*'diff(y(t),t)-cc*y(t); [[feedback:Result]]

]]> \({@n1@} \frac{d^2y}{dt^2} - {@bc@} \frac{dy}{dt} - {@cc@} y = 0\) gives \(y = A{@e^(m1*t)@}+B{@e^(m2*t)@}\) 1 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> none 1 i cos-1 [ ans1 algebraic correctanswer 20 1 0 0 solve 1 1 1 1 1 Result 1.0000000 1 p:ev(q,y(t)=ans1,nouns,fullratsimp); l:setify(listofvars(ans1)); l:setdifference(l,set(t)); l:listify(l); lv:length(l); b1:ev(ans1,t=0,fullratsimp); b2:ev(ans1,t=1,fullratsimp); m:if not(b2=0) then fullratsimp(b1/b2) else 0; m:float(m); p2:ev(q,y(t)=subst(t,x,ans1),nouns,fullratsimp); 0 AlgEquiv p 0 0 = 1.0000000 1 Sat-ODE = 0.0000000 3 Does-Not-Sat-ODE Your answer should satisfy the differential equation, but in fact when we substitute your expression into the differential equation we get \[{@p@}\] which is not zero, so you must have done something wrong.

]]> 1 AlgEquiv lv 2 0 + 0.0000000 2 Has-2-consts = 0.7500000 -1 Wrong-#-consts You should have a general solution, which includes unknown constants. Your answer satisfies the differential equation, but does not have the correct number of unknown constants.

]]> 2 AlgEquiv numberp(m) true 0 = 0.5000000 -1 Sol-not-lin-ind Your general solution should be a sum of two linearly independent components, but is not.

]]> = 1.0000000 -1 Correct 3 AlgEquiv p2 0 0 + 0.0000000 -1 Result-4-T Actually, if we make the substitution \(x=t\) then it looks like your answer does satisfy the equation. You have probably used the wrong variable in your answer!

]]> - 0.0000000 -1 Result-4-F 258189496 1454874554 618666105 1089015490 976055893 1539806997 141387721 1331542353 1497396135 410647422 1 ans1 correctanswer Result 1.0000000 0.0000000 Correct 2 ans1 correctanswer-c Result 0.0000000 0.0000000 Result-4-F 3 ans1 altanswer Result 1.0000000 0.0000000 Correct 4 ans1 wronganswer2 Result 0.7500000 0.0000000 Wrong-#-consts 5 ans1 A*e^(m2*t)+B*e^(m2*t) Result 0.5000000 0.0000000 Sol-not-lin-ind 6 ans1 ev(subst(x,t,correctanswer),simp) Result 0.0000000 0.0000000 Result-4-T 2nd-Order-ODE-2-undamped-complex Solve \[{@n1@} \frac{d^2y}{dt^2} + {@cc@} y = 0.\]
\(y(t)\) = [[input:ans1]] [[validation:ans1]]

]]> To solve the differential equation \[{@n1@} \frac{d^2y}{dt^2} + {@cc@} y = 0\] Let \(y = e^{mt}\) and hence \(\frac{dy}{dt}=me^{mt}\) and \(\frac{d^2y}{dt^2} = m^2e^{mt}\)
Substituting for \(y\) gives \[{@n1@}m^2e^{mt} + {@cc@}e^{mt} = 0\] and dividing throughout by \(e^{mt}\) and {@n1@} gives \[m^2 + {@cc/n1@} = 0\] which can be solved to give {@m@}
For roots of \(m\) in the form \(a±bi\), the solution to the differential equation is of the form \(y = e^{at}(A\cos(bt)+B\sin(bt))\).
Therefore \(y=A\cos({@bvar@}t)+B\sin({@bvar@}t)\).

]]> 1.0000000 0.0000000 0 n1:rand(8)+2; n2:rand(7)+1; cc:n1*(n2^2); m:solve(n1*m^2 + cc = 0,m); m1:rhs(m[1]); m2:rhs(m[2]); avar:realpart(m1); bvar:abs(imagpart(m1)); correctanswer:e^(avar*t)*(A*cos(bvar*t)+B*sin(bvar*t)); altanswer:e^(avar*t)*(B*cos(bvar*t)+A*sin(bvar*t)); wronganswer1:e^(avar*t)*(Q*cos(bvar*t)+D*sin(bvar*t)); wronganswer2:e^(avar*t)*B*cos(bvar*t); wronganswer3:e^(avar*t)*A*sin(bvar*t); q:n1*'diff(y(t),t,2)+cc*y(t); [[feedback:Result]]

]]> \({@n1@} \frac{d^2y}{dt^2} + {@cc@} y = 0\) gives \( A\cos({@bvar@}t)+B\sin({@bvar@}t)\) \({@q@}\) 1 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> none 1 i cos-1 [ ans1 algebraic correctanswer 20 1 0 0 1 1 1 1 1 Result 2.0000000 1 p:ev(q,y(t)=ans1,nouns,fullratsimp); l:setify(listofvars(ans1)); l:setdifference(l,set(t)); l:listify(l); lv:length(l); b1:ev(ans1,t=0,fullratsimp); b2:ev(ans1,t=1,fullratsimp); m:if not(b2=0) then fullratsimp(b1/b2) else 0; m:float(m); 0 AlgEquiv p 0 0 = 1.0000000 1 Result-0-T = 0.0000000 -1 Result-0-F Your answer should satisfy the differential equation, but in fact when we substitute your expression into the differential equation we get \[{@p@}\] which is not zero, so you must have done something wrong.

]]> 1 AlgEquiv lv 2 0 = 1.0000000 2 Result-1-T = 0.7500000 -1 Result-1-F You should have a general solution, which includes unknown constants. Your answer satisfies the differential equation, but does not have the correct number of unknown constants.

]]> 2 AlgEquiv numberp(m) true 0 + 0.0000000 -1 Result-2-T Your general solution should be a sum of two linearly independent components, but is not.

]]> + 0.0000000 -1 Result-2-F 977553297 817649043 1841797587 206612153 1928692521 108955355 1654432021 1607577411 559509500 1391994445 1 ans1 correctanswer Result 1.0000000 0.0000000 Result-2-F 2 ans1 correctanswer-c Result 0.0000000 0.0000000 Result-0-F 3 ans1 altanswer Result 1.0000000 0.0000000 Result-2-F 4 ans1 r*sin(n2*t+rho) Result 1.0000000 0.0000000 Result-2-F 5 ans1 wronganswer2 Result 0.7500000 0.0000000 Result-1-F 6 ans1 wronganswer3 Result 0.7500000 0.0000000 Result-1-F 7 ans1 A*e^(i*n2*t)+B*e^(-i*n2*t) Result 1.0000000 0.0000000 Result-2-F 2nd-Order-ODE-3-real-repeated-BC Find the general solution of \[{@n1@} \frac{\mathrm{d}^2y}{\mathrm{d}t^2} + {@n1*n2^2@} y ={@2*n1*n2@} \frac{\mathrm{d}y}{\mathrm{d}t} .\]
\(y(t)\) = [[input:ans1]] [[validation:ans1]]

[[feedback:Result]]

Find the particular solution subject to \(y(0)=\){@ev(ca,t=0)@} and \( y'(0)=\){@ev(cad,t=0)@}.
\(y(t)\) = [[input:ans2]] [[validation:ans2]]

[[feedback:Result2]]

]]> To solve the differential equation \[{@n1@} \frac{\mathrm{d}^2y}{\mathrm{d}t^2} + {@n1*n2^2@} y ={@2*n1*n2@} \frac{\mathrm{d}y}{\mathrm{dt}} .\]
first write this in standard form \[\frac{\mathrm{d}^2y}{\mathrm{d}t^2} - {@2*n2@} \frac{\mathrm{d}y}{\mathrm{d}t}+ {@n2^2@} y =0 .\]
Let \(y = e^{mt}\) and hence \(\frac{dy}{dt}=me^{mt}\) and \(\frac{d^2y}{dt^2} = m^2e^{mt}\).
Substituting for \(y\) gives \[m^2e^{mt} -{@2*n2@}e^{mt}+ {@n2^2@}me^{mt}= 0\] and dividing throughout by \(e^{mt}\) gives \[m^2 -{@2*n2@}m+{@n2^2@}=(m-{@n2@})^2=0\] which can be solved to give a repeated root \(m={@n2@}\). The solution is then of the form \(y = A{@e^(n2*t)@}+Bt{@e^(n2*t)@}\).

For the second part we evaluate \(y(0)\) and \(y'(0)\) and use the specified information to give

\[ y(0) = {@ev(correctanswer,t=0)@} = {@ev(ca,t=0)@} \]
\[ y'(t) = {@factor(diff(correctanswer,t))@}\]
so that
\[ y'(0) = {@ev(dcorrectanswer,t=0)@} = {@ev(cad,t=0)@}.\]
Solving these equations for \(A\) and \(B\) gives
\[ A={@AA@}, \quad B={@BB@}\]
so that that solution is \(y(t)={@ca@}\).

]]> 4.0000000 0.0000000 0 n1:rand(8)+2 n2:rand(3)+2 q:n1*'diff(y(t),t,2)-2*n1*n2*'diff(y(t),t)+n1*n2^2*y(t) correctanswer:A*e^(n2*t)+B*t*e^(n2*t) dcorrectanswer:diff(correctanswer,t) AA:rand(2)+2 BB:rand(2)+2 ca:ev(correctanswer,[A=AA,B=BB]) cad:diff(ca,t) \({@n1@} \frac{\mathrm{d}^2y}{\mathrm{d}t^2} + {@n1*n2^2@} y ={@2*n1*n2@} \frac{\mathrm{d}y}{\mathrm{d}t}\) gives \(y = A{@e^(n2*t)@}+Bt{@e^(n2*t)@}\) and {@ca@} 1 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> none 1 i cos-1 [ ans1 algebraic correctanswer 25 1 0 0 solve 1 1 1 1 1 ans2 algebraic ca 25 1 0 0 1 0 0 1 1 Result 2.0000000 1 p:ev(q,y(t)=ans1,nouns,fullratsimp); l:setify(listofvars(ans1)); l:setdifference(l,set(t)); l:listify(l); lv:length(l); b1:ev(ans1,t=0,fullratsimp); b2:ev(ans1,t=1,fullratsimp); m:if not(b2=0) then fullratsimp(b1/b2) else 0; m:float(m); 0 AlgEquiv p 0 0 = 1.0000000 1 Sat-ODE = 0.0000000 -1 Does-Not-Sat-ODE Your answer should satisfy the differential equation, but in fact when we substitute your expression into the differential equation we get \[{@p@}\] which is not zero, so you must have done something wrong.

]]> 1 AlgEquiv lv 2 0 + 0.0000000 2 Has-2-consts = 0.5000000 -1 Wrong-#-consts You should have a general solution, which includes unknown constants. Your answer satisfies the differential equation, but does not have the correct number of unknown constants.

]]> 2 AlgEquiv numberp(m) true 0 = 0.5000000 -1 Sol-not-lin-ind Your general solution should be a sum of two linearly independent components, but is not.

]]> = 1.0000000 -1 Correct Result2 2.0000000 1 dans2:diff(ans2,t) 0 AlgEquiv ev(ans2,t=0) ev(ca,t=0) 0 = 0.5000000 1 BC-y(0)-ok = 0.0000000 1 BC-y(0)-not-ok Your answer should satisfy \(y(0)=\){@ev(ca,t=0)@}. In fact, evaluating {@ans2@} at \(t=0\) gives {@ev(ans2,t=0)@}.

]]> 1 AlgEquiv ev(dans2,t=0) ev(cad,t=0) 0 + 0.5000000 -1 BC-y'(0)-ok - 0.0000000 -1 BC-y'(0)-not-ok Your answer should satisfy \(y'(0)=\){@ev(cad,t=0)@}. In fact, evaluating \(\frac{\mathrm{d}y}{\mathrm{dt}}\left({@ans2@}\right)\) at \(t=0\) gives {@ev(dans2,t=0)@}.

]]> 1159279826 1269312969 738005696 193204862 48103118 1863964533 112335941 1 ans1 correctanswer ans2 ca Result 1.0000000 0.0000000 Correct Result2 1.0000000 0.0000000 BC-y'(0)-ok 2 ans1 correctanswer-c Result 0.0000000 0.0000000 Does-Not-Sat-ODE 3 ans1 A*e^(n2*t)+B*e^(n2*t) ans2 Result 0.5000000 0.0000000 Sol-not-lin-ind 4 ans1 t*%e^(n*2*t)*B+%e^(n*2*t)*A ans2 ev(ca+t*%e^(n2*t),simp) Result 0.0000000 0.0000000 Does-Not-Sat-ODE Result2 0.5000000 0.0000000 BC-y'(0)-not-ok 5 ans1 A*e^(n2*t) ans2 Result 0.5000000 0.0000000 Wrong-#-consts 2nd-Order-ODE-find-parameter Consider the following ODE \[\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + {@n1@} \frac{\mathrm{d}y}{\mathrm{d}t} + k^2 y(t) =0.\]

For what values of \(k>0\) does the graph of the solution look qualitatively of the form shown below?
{@plot(qa3,[x,0,100])@}

[Notes, (1) you are not being asked to solve the ODE and (2) "qualitative" means of the same general shape.]

You must enter your answer as inequalities in \(k\) to represent a range of real numbers. For example,

to represent the interval \( [2,3) \) you might type 2<=k and k<3
to represent the interval \( [0,\infty) \) you might type 0<=k
you can connect individual inequalities using and or or, as the logic dictates in your answer.

[[input:ans1]]

[[validation:ans1]]

]]> Consider the ODE \[\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + {@n1@} \frac{\mathrm{d}y}{\mathrm{d}t} + k^2 y(t) =0.\]

Substitute in \(y(t)=e^{mt}\) so we have \[m^2 + {@n1@} m + k^2 m =0.\]
The solutions to this equation are given by \[ m = \frac{-{@n1@} \pm \sqrt{{@n1@}^2-4\times k^2}}{2} = -{@n1/2@}\pm \sqrt{{@n1^2/4@}-k^2}\]
The only way to have oscillatory solutions is to have a complex root, and this can only happen if \({@n1^2/4@}-k^2<0\), i.e. if \(k^2>{@n1^2/4@}\). Since we know \(k>0\) we can eliminate the negative values to give \(k>\sqrt{{@n1^2/4@}}\), i.e. \(k>{@n1/2@}.\).

]]> 1.0000000 0.1000000 0 n1/2 q:'diff(y,x,2) + n1*'diff(y,x)+(sqrt(n1/2)+2)*y = 0; qa:ode2(q,y,x); qa2:ic2(qa,x=0,y=1,'diff(y,x)=0) qa3:rhs(qa2)]]> [[feedback:prt1]]

]]> {@ta@} 1 0 0 Correct answer, well done.

]]> Your answer is partially correct.

]]> Incorrect answer.

]]> none 1 i cos-1 [ ans1 algebraic ta 15 1 0 0 1 0 1 1 1 prt1 1.0000000 1 0 AlgEquiv ans1 ta 0 = 1.0000000 -1 prt1-1-T = 0.0000000 1 prt1-1-F 1 AlgEquiv ans1 =sqrt(n1/2)]]> 1 + 0.5000000 -1 prt1-2-T - 0.0000000 2 prt1-2-F 2 AlgEquiv ans1 1 + 0.0000000 -1 prt1-3-T - 0.0000000 3 prt1-3-F 3 AlgEquiv ans1 1 + 0.0000000 -1 prt1-4-T - 0.0000000 4 prt1-4-F 4 AlgEquiv ans1 n1/2]]> 1 + 0.5000000 -1 prt1-5-T This is basically correct, but you need to write your inequality in simplified form.

]]> - 0.0000000 5 prt1-5-F 5 AlgEquiv ans1 sqrt(n1/2) or k<-sqrt(n1/2)]]> 1 + 0.5000000 -1 prt1-6-T You can eliminate the negative values of \(k\) using the given condition that \(k>0\).

]]> - 0.0000000 -1 prt1-6-F 1685192574 1005426202 371097510 1456215945 1 ans1 ta prt1 1.0000000 0.0000000 prt1-1-T 2 ans1 =sqrt(n1/2)]]> prt1 0.5000000 0.1000000 prt1-2-T 3 ans1 prt1 0.0000000 0.1000000 prt1-3-T 4 ans1 n1/2]]> prt1 0.5000000 0.1000000 prt1-5-T 5 ans1 =n1/2]]> prt1 0.0000000 0.1000000 prt1-6-F Is this separable The differential equation \[ \frac{dy}{dx} = x+y\] is separable.

]]> 1.0000000 1.0000000 0 true false Linearity If \(y_1(x)\) and \(y_2(x)\) are both solutions of
\[ \frac{\mathrm{d}y}{\mathrm{d}x}=g(y,x)\]
then for any numbers \(c_1\) and \(c_2\) the function \(x\mapsto c_1y_1(x)+c_2y_2(x)\) is also a solution of the differential equation.

]]> 1.0000000 1.0000000 0 true false