Facts needed for these DE Polls
- The solution of the initial value problem for $y^{(n)}+p_1(t) y^{(n-1)}+\dots+p_n(t) y=g(t)$ exists on the interval of continuity of all functions that contains $t_0$.
- Wronskian $W[y_1,y_2,y_3]=\det \begin{bmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \\ \end{bmatrix}$ is used to determine if we found a fundamental set of solutions.
- The general solution of homogeneous constant coefficients equation $a_0 y^{(n)}+a_1 y^{(n-1)}+\dots+a_n y=0$ is determined by the roots of the characteristic equation $a_0r^n+a_1 r^{n-1}+\dots+a_{n-1}r+a_n =0$. WolframAlpha and ODESolver work well on such problems.
The solution of the initial value problem $(t^2-1)y'''+y'+y=\frac{1}{\sin t}$, with $y(1/2)=0,y'(1/2)=0,y''(1/2)=0$ is sure to exist on the interval
- $(-\pi,\pi)$
- $(-1,1)$
- $(-1,0)$
- $(0,1)$
- none of the above
Answer: (D)
Solution The discontinuity points, in increasing order, are $\dots,-\pi,-1,0,1,\pi,2\pi,\dots$. Where is $t_0$? In $(0,1)$.
The solution of the initial value problem $(t^2-1)y'''+y'+y=\frac{1}{\sin t}$, with $y(2)=0,y'(2)=0,y''(2)=0$ is sure to exist on the interval
- $(-\pi,\pi)$
- $(-1,1)$
- $(-1,0)$
- $(0,1)$
- none of the above
Answer: (E)
Solution The discontinuity points, in increasing order, are $\dots,-\pi,-1,0,1,\pi,2\pi,\dots$. Where is $t_0$? In $(1,\pi)$.
The solution of the initial value problem $(4-t^2)y'''+y'+(4-t^2)y=(4-t^2)\tan t$, with $y(1.6)=0, y'(1.6)=0, y''(1.6)=0$ is sure to exist on the interval
- $(-\pi/2,\pi/2)$
- $(-2,-\pi/2)$
- $(\pi/2,2)$
- $(-\infty,-2)$
- none of the above
Answer: (C)
Solution The discontinuity points, in increasing order, are $\dots-\frac{3}{2}\pi,-2,-\pi/2,\pi/2,2,\pi,\frac{3}{2}\pi,\dots$. Where is $t_0$? In $(\pi/2,2)$.
The solution of the initial value problem $(4-t^2)y'''+y'+(4-t^2)y=(4-t^2)\tan t$, with $y(0)=0, y'(0)=0, y''(0)=0$ is sure to exist on the interval
- $(-\pi/2,\pi/2)$
- $(-2,-\pi/2)$
- $(-\pi/2,2)$
- $(-\infty,-2)$
- none of the above
Answer: (A)
Solution
The discontinuity points, in increasing order, are $\dots-\frac{3}{2}\pi,-2,-\pi/2,\pi/2,2,\pi,\frac{3}{2}\pi,\dots$. Where is $t_0$? In $(-\pi/2,\pi/2)$.
Compute the Wronskian $W[y_1,y_2,y_3]$ with $y_1=t, y_2=t^2,y_3=t^3$
Answer: $2t^3$
Solution
$$W[t,t^2,t^3]=\det\begin{bmatrix} t & t^2 & t^3 \\ 1 & 2t & 3t^2 \\ 0& 2 & 6t \\ \end{bmatrix}=t \det \begin{bmatrix} 2 t & 3 t^2 \\ 2 & 6t\end{bmatrix} -t^2\det\begin{bmatrix}1 & 3 t^2 \\ 0 & 6t\end{bmatrix}+t^3 \det \begin{bmatrix}1 & 2t \\ 0 &2\end{bmatrix}= 2t^3$$
Compute the Wronskian $W[y_1,y_2,y_3]$ with $y_1=t, y_2=t^2,y_3=1/t$
Answer: $W=6/t$
Solution $W[t,t^2,\frac{1}{t}]=\det\begin{bmatrix} t & t^2 & 1/t \\ 1 & 2t & -1/t^2 \\ 0& 2 & 2/t^3 \\ \end{bmatrix}=\frac{6}{t}$
To find the general solution of $y''- y'- 2 y=10 \cos t $ a student a student seeks a particular solution of the form $y_*=A \cos t+B\sin t$.What value of $A$ will she get?
Answer: $A=-3$
Solution The general solution is $y=C_1e^{-t}+C_2e^{2t}+y_*$, so $y_*=A \cos t+B\sin t$ is an appropriate guess. We now compute $y_*'=-A\sin t+B\cos t$, $y_*''=-A\cos t-B\sin t$ and put them into the equation. We get $$L[y_*]= -A\cos t-B\sin t+A\sin t -B\cos t - 2 A \cos t- 2B \sin t =-(3A+B)\cos t+(A-3B)\sin t$$ Since $L[y_*]=10\cos t$ we get $3A+B=-10$ and $(A-3B)=0$ so $A=-3$ and $B=-1$. The general solution is $$ y=C_1e^{-t}+C_2e^{2t}-3 \cos t- \sin t $$ This answer can also be read out from WolframAlpha
To find the solution of $y''- y'- 2 y=10 \cos t $ with the initial value $y(0)=2,y'(0)=3$ a student a student seeks a particular solution of the form $y_*=A \cos t+B\sin t$. After some computation she determines the undetermend coefficients and finds a particular solution $y_*=-3 \cos t- \sin t$ She writes the general solution as $$y=C_1e^{-t}+C_2e^{2t}-3 \cos t- \sin t $$ and proceeds to compute the constants $C_1,C_2$.What value of $C_1$ will she get for the solution of the initial value problem?
Answer: $C_1=2$
Solution According to WolframAlpha, the solution is $y=2 e^{-t} + 3 e^{2 t} - 3 \cos(t) - \sin(t)$.
Solve the initial value problem $y^{iv}+4 y''+ 4 y=0,\;y(0)= 4, y'(0)= 2, y''(0)= 4, y'''(0)= 4$.Compute the decimal value of $y(1)$.
Answer: $y(1)=7.29645$
Solution According to WolframAlpha, $y(t)= 3 \sqrt{2} t \sin \left(\sqrt{2} t\right)+2 \sqrt{2} \sin \left(\sqrt{2} t\right)-2 t \cos \left(\sqrt{2} t\right)+4 \cos \left(\sqrt{2} t\right)$. With enough patience, this answer can also be computed by hand.
Solve the initial value problem $y^{iv}+2 y''+ y=0,\;y(0)= 0, y'(0)= 2, y''(0)=0, y'''(0)= 2$.Compute the decimal value of $y(1)$.
Answer: $y(1)=2.28528$
Solution According to WolframAlpha, $y(t)= 4 \sin (t)-2 t \cos (t)$. With enough patience, this answer can also be computed by hand.
Solve the initial value problem $y^{iv}+2 y''+ y=0,\;y(0)= 0, y'(0)= 2, y''(0)=0, y'''(0)= 2$.Compute the decimal value of $y(1)$.
Answer: $y(1)=2.76355$
Solution According to WolframAlpha, $y(t)=2 t \sin (t)+2 \cos (t)$. With enough patience, this answer can also be computed by hand.
Solve the initial value problem $y'''+ y=0,\;y(0)= 6, y'(0)= 0, y''(0)=0$.Compute the decimal value of $y(1)$.
Answer: $y(1)=5.00832$
Solution According to WolframAlpha, $y(t)=2 e^{-t}+4 e^{t/2} \cos \left(\frac{\sqrt{3} t}{2}\right)$. With enough patience, this answer can also be computed by hand.
Money Type D:
A college graduate borrows 50,000 to buy a Tesla S at an interest rate of 6%. Anticipating steady salary increases, he expects to steady increase his monthly payments, with $m(t)=500+10 t$ after $t$ months.Assuming that this payment schedule can be maintained, what will be the amount of loan in 5 years? Will the loan be fully paid in 6 years?
SolutionLets use months as units of $t$. The loan amount changes at the rate $B'= \frac{.06}{12} B - (500+ 10 t)$ with the initial condition $B(0)=50000$. This is a linear but non-separable equation. Solution from WolframAlpha is $B(t)=500000. - 450000. e^{0.005 t} + 2000. t$. Answer: $B(60)=12563.5$ (From the graph of $B(t)$, the loan will be paid of in about 71.17 months.)
Type D: hard
A tank originally contains 100 gal of fresh water and has capacity of 200 gal. Then water containing ${\frac 12}$ lb of salt per gallon is poured into the tank at a rate of 3 gal/min, and the mixture is allowed to leave at a rate of 2 gal/min. Setup the differential equation for the amount $A(t)$ of salt in the tank at time $t$. What is the amount of salt in the tank at the time of overespil?l
Solution Water level increases at the rate of 1 gal/min, so the volume of water at time $t$ is $V(t)=100+t$. The water with overspill at $t=100$ min.
The "rate in - rate out" formula gives $A'(t)=3\times \frac12 -2\times \frac{A}{100+t}$ and $A(0)=0$ (fresh water initialy). Solving the equation by WolframAlpha we get $A(t)=\frac{t (30000 + 300 t + t^2)}{2 (100 + t)^2}$ and $A(100)=\frac{175}{2}=87.5 $