In the power solution $y=\sum_{n=0}^\infty a_n (x-\tfrac12)^n$ of the initial value problem $(1-x^2)y'''+y'+y=\sin (\pi x)$, with $y(1/2)=1,y'(1/2)=-1,y''(1/2)=1$, determine the value of $a_3$. Write your answer as a fraction using a slash like 2/3.
Answer: $a_3=2/9$
Solution $y(x)=a_0+a_1(x-\tfrac12) +a_2(x-\tfrac12)^2 +a_3(x-\tfrac12)^2+\dots$. On the other hand, by Taylor's formula, $y(x)=y(1/2)+ y'(1/2)(x-\tfrac12)+\frac{ y''(1/2)}{2!}(x-\tfrac12)^2+\frac{ y'''(1/2)}{3!}(x-\tfrac12)^3+\dots$. The first three values are given, and we only need to compute $a_3=\frac{ y'''(1/2)}{3!}$. To computer $y'''$ we use the differential equation $$y'''(1/2)=\frac{\sin(\pi/2)-y(1/2)-y'(1/2)}{1-\tfrac14}=\frac{1-1+1}{3/4}=4/3$$ so $a_3=2/9$
In the power solution $y=\sum_{n=0}^\infty a_n x^n$ of the initial value problem $\cos (x)y''+y'+y=\sin (\pi x)$, with $y(0)=12,y'(0)=-24 $, determine the value of $a_2$.
Answer: $a_2=6$
Solution $y(x)=a_0+a_1 x+a_2 x^2+\dots$. On the other hand, by Taylor's formula, $y(x)=y(0)+y'(0) x+ \frac{y''(0)}{2}x^2+\dots$. So using the differential equation we compute $a_2=\tfrac12 y''(0)=\frac{\sin 0-y(0)-y'(0)}{2\cos 0}=\tfrac12 (24-12)=6$
In the power solution $y=\sum_{n=0}^\infty a_n x^n$ of the initial value problem $ y''+2 y'+ 3 y=2 e^x$, with $y(0)=2,y'(0)=-4 $, determine the value of $a_2$.
Answer: $a_2=2$
Solution $y(x)=a_0+a_1 x+a_2 x^2+\dots=y(0)+y'(0) x+ \frac{y''(0)}{2}x^2+\dots$. So $a_2=\tfrac12y''(0)=\frac{2e^0-3 y(0)- 2y'(0)}{2}=\frac{2-6+8}{2}=2$
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 $(-\pi/2,\pi/2)$.
What is the degree of the polynomial $$\frac{d}{dx} \sum_{n=3}^{77}\frac{x^n}{n^2+1}$$
Answer: $76$
Solution
$\frac{d}{dx} \sum_{n=3}^{77}\frac{x^n}{n^2+1}=\sum_{n=3}^{77}\frac{n}{n^2+1}x^{n-1}=\frac{3}{10}x^2+\frac{5}{17}x^3+\dots+\frac{77}{77^2+1}x^{76}$
What is the lowest power of $x$ after you simplify the polynomial $$ \frac{d}{dx}\left( \sum_{n=3}^{77}\frac{x^n}{n^2+1}\right)-\frac{1}{10}\sum_{k=3}^{10} k x^{k-1} $$
Answer: $3$
Solution $ \frac{d}{dx} \sum_{n=3}^{77}\frac{x^n}{n^2+1}-\frac{1}{10}\sum_{k=3}^{10} k x^{k-1}=\frac{d}{dx} \left(x^3/10+x^4/17+\dots\right)- \frac{1}{10}\left(3x^2+4 x^3+\dots\right)= (\frac{3}{10}-\frac{3}{10})x^2 +(\frac{4}{17}-\frac{4}{10})x^3+\dots$, so the lowest power is $x^3$ with the coefficient $-\frac{14}{85}$
Simplify polynomial $p(x)=\sum_{k=2}^{15} k^4 x^{k-1}-\sum_{n=3}^{17} n^2 x^{n+1}$ by writing is as $\sum a_j x^j$. Determine the value of $a_5$
Answer: $1280$
Solution A brute-force calculation gives $p(x)=-289 x^{18}-256 x^{17}-225 x^{16}-196 x^{15}+50456 x^{14}+38272 x^{13}+28440 x^{12}+20636 x^{11}+14560 x^{10}+9936 x^9+6512 x^8+4060 x^7+2376 x^6+1280 x^5+616 x^4+256 x^3+81 x^2+16 x$. (There are simpler ways to solve this.)
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 $y=2 e^{-t} + 3 e^{2 t} - 3 \cos(t) - \sin(t)$.
With $y(x)=\sum_{n=0}^\infty a_n x^n$ rewrite $x y(x)+y'(x)$ as a power series. What coefficient do we get at $x^{75}$?
- $75 a_{75}+a_{74}$
- $a_{75}+75 a_{74}$
- $76 a_{75}+ a_{74}$
- $75 a_{74}+ a_{76}$
- None of the above
Answer: E
Solution The expression is $x \sum_{n=0}^\infty a_n x^n+\frac{d}{dx}\sum_{n=0}^\infty a_n x^n =\sum_{n=0}^\infty a_n x^{n+1}+\sum_{n=0}^\infty n a_n x^{n-1}$. So at $x^{75}$ we get $a_{74}+76 a_{76}$
With $y(x)=\sum_{n=0}^\infty a_n x^n$, rewrite, $y(x)+x y'(x)$ as a power series. What coefficient do we get at $x^{75}$?
- $75 a_{75}+a_{74}$
- $a_{75}+75 a_{74}$
- $76 a_{75}+ a_{74}$
- $75 a_{74}+ a_{76}$
- None of the above
Answer: E
Solution The expression is $\sum_{n=0}^\infty a_n x^n+x \frac{d}{dx}\sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty a_n x^{n}+\sum_{n=0}^\infty n a_n x^{n}$. So at $x^{75}$ we get $76 a_{75}$
With $y(x)=\sum_{n=0}^\infty a_n x^n$, rewrite, $xy(x)+x y'(x)$ as a power series. What coefficient do we get at $x^{75}$?
- $75 a_{75}+a_{74}$
- $a_{75}+75 a_{74}$
- $76 a_{75}+ a_{74}$
- $75 a_{74}+ a_{76}$
- None of the above
Answer: (A)
Solution The expression is $x\sum_{n=0}^\infty a_n x^n+x \frac{d}{dx}\sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty a_n x^{n+1}+\sum_{n=0}^\infty n a_n x^{n}$. So at $x^{75}$ we get $a_{74}+75 a_{75}$
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 $