Trig identity of use here is $\cos(\alpha-\beta)=\cos\alpha\cos \beta+\sin\alpha\sin\beta$
According to WolframAlpha, the solution of the initial value problems $u''+4u=0$, $u(0)=2$, $u'(0)=2$ is $$u=2\cos(2t)+\sin (2t)$$ Determine the amplitude and the phase of this motion.
Hint: Write this answer in the form $u=R\cos(2t-\delta)$ and write the decimal values of $R$ and $\delta$.
Answer: $R=\sqrt{5}\approx 2.23607$, and $\delta=\arctan(1/2)\approx 0.4663$
Solution
$u=\sqrt{5}\left(\frac{2}{\sqrt{5}}\cos(2t)+\frac{1}{\sqrt{5}}\sin (2t)\right)$ and $\frac{1}{\sqrt{5}}=\sin \delta$ when $\delta=\arctan(1/2)$
According to WolframAlpha, the solution of the initial value problems $u''+4u=0$, $u(0)=1$, $u'(0)=6$ is $$u=\cos(2t)+3\sin (2t)$$ Determine the amplitude and the phase of this motion.
Hint: Write this answer in the form $u=R\cos(2t-\delta)$ and write the decimal values of $R$ and $\delta$.
Answer: $R=\sqrt{10}\approx 3.16228$ $\delta= \arctan(3)\approx 1.24905$
Solution (see previous problem)
Determine the amplitude of the oscillations described by equation $u''+u=0,\; u(0)=3,u'(0)=4$.
Answer: $R=5$
Solution
The general solution of the homogeneous equation is $u=C_1 \cos t+C_2 \sin t$, and the initial value is solved by $u=3\cos t+4 \sin t$. So the amplitude is $R=\sqrt{3^2+4^2}=5$.
To solve the initial value problem $y''-4 y'+ 4 y=18 e^{2t},\; y(0)=2,y'(0)=0$ a student seeks a particular solution of the form $y_*=A t e^{2t}$.What value of $A$ will she get?
Answer: none
Solution The general solution of the homogeneous equation is $y=C_1 e^{2t}+C_2 t e^{2t}$, so her guess will not work at all as $L[A t e^{2t}]=0$ cannot match $18e^{2t}$. The proper guess for a double root is $y_*=A t^2 e^{2t}$!
To solve the initial value problem $y''-4 y'+ 4 y=18 e^{2t},\; y(0)=2,y'(0)=0$ a student seeks a particular solution of the form $y_*=A t^2 e^{2t}$.What value of $A$ will she get?
Answer: $A=9$
Solution
One can solve this problem by hand, or you can read this answer from WolframAlpha
To solve the initial value problem $y''-4 y'+ 4 y=18 e^{2t},\; y(0)=2,y'(0)=0$ a student seeks a particular solution of the form $y_*=A t^2 e^{2t}$ and finds out that $A=9$. She now want to determine the answer using the general solution $y=C_1 e^{2t}+C_2 t e^{2t}+y_*$.What value of $C_1$ will she get?
Answer: $C_1=2$
Solution The solution is $y=e^{2t}(9 t^2-4t+2)$.
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 $y=2 e^{-t} + 3 e^{2 t} - 3 \cos(t) - \sin(t)$.
To solve the initial value problem $y''+4 y'+ 4 y=8 t^2,\; y(0)=2,y'(0)=0$ a student seeks a particular solution of the form $y_*=A t^2$.What value of $A$ will she get?
Answer: none
Solution $L[A t^2]=2A+8At +4At^2$ cannot be equal to $8t^2$ no matter what we do to $A$. (Remember that $A$ is a constant, not a function of $t$)
To solve the initial value problem $y''+4 y'+ 4 y=8 t^2,\; y(0)=2,y'(0)=0$ a student seeks a particular solution of the form $y_*=A t^2$.What value of $A$ will she get?
Answer: none
Solution $L[A t^2]=2A+8At +4At^2$ cannot be equal to $8t^2$ no matter what we do to $A$. (Remember that $A$ is a constant, not a function of $t$)
To solve the initial value problem $y''+4 y'+ 4 y=8 t^2,\; y(0)=2,y'(0)=0$ a student seeks a particular solution of the form $y_*=A t^2+Bt+C$.What value of $A$ will she get?
Answer: $A=2$
Solution According to WolframAlpha the general solution is $y=C_1 e^{-2t}+C_2 t e^{-2t}+2t^2-4t+3$
To solve the initial value problem $y''+4 y'+ 4 y=8 t^2,\; y(0)=2,y'(0)=0$ a student seeks a particular solution of the form $y_*=A t^2+Bt+C$. After some work he determines that $y_*=+2t^2-4t+3$ so the general solution is $y=C_1 e^{-2t}+C_2 t e^{-2t}+2t^2-4t+3$What value of $C_2$ will he get?
Answer: $C_2=2$
Solution
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 $