Facts needed for these DE Polls
- $a+ i b=Re^{i\theta}$ with $R\geq 0$ and $-\pi<\theta\leq \pi$ (The book sometimes asks for $0\leq \theta<2\pi$)
In the representation $1+3i=Re^{i\theta}$, what is the value of $R$? (Compute the numerical value.)
Answer: $R=\sqrt{10}\approx 3.16228$
Solution $1+3i=R(\cos \theta+ i \sin \theta)$ so $R^2=(R \cos \theta)^2+(R\sin \theta)^2=1+9=10$
In the representation $1+3i=Re^{i\theta}$, what is the value of $\theta$?. (Compute the numerical value.)
Answer: $\theta=1.24905$
Solution $1+3i=R(\cos \theta+ i \sin \theta)$ so $\tan \theta=3$.
In the representation $1-2i=Re^{i\theta}$, what is the value of $R$? (Compute the numerical value.)
Answer: $R=2.23607$
Solution $R=\sqrt{5}$.
In the representation $1-2i=Re^{i\theta}$, what is the value of $\theta$?. (Compute the numerical value.)
Answer: $\theta=-1.10715$
Solution $\tan \theta=-2$ so $\theta=-\arctan(2)$.
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 $