Determine the recurrence for the coefficients of the power series solution $y=\sum_{n=0}^\infty a_n x^n$ for equation $(1-x)y'' +x y'+y=0$, valid for large enough $n$.
- $$a_{n+2}=-\frac{1}{(1-x)(n+2)}a_n$$
- $$a_{n+2}=-\frac{n}{n+2}a_n$$
- $$a_{n+2}=-\frac{n}{n+2} a_{n+1}+\frac{1}{n+2}a_n$$
- $$a_{n+2}=-\frac{1+nx}{(1-x)(n+2)(n+1)}a_n $$
- None of the above
Answer: (C)
Solution $y=\sum_{n=0}^\infty a_n x^n$, $$y'=\sum_{n=0}^\infty (n+1)a_{n+1}x^n$$ so $x y'=\sum_{n=1}^\infty na_{n}x^n$ $$y''=\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n$$ so $xy''=\sum_{n=1}^\infty (n+1)na_{n+1}x^n$ \begin{multline*} L[y]=(1-x)y'' +x y'+y \\=\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n -\sum_{n=1}^\infty (n+1)na_{n+1}x^n+\sum_{n=1}^\infty na_{n}x^n+\sum_{n=0}^\infty a_n x^n \\ = 2a_2-a_0 +\sum_{n=1}^\infty \left((n+2)(n+1)a_{n+2}-(n+1)n a_{n+1}+(n+1)a_n\right)x^n \end{multline*}
Determine the recurrence for the coefficients of the power series solution $y=\sum_{n=0}^\infty a_n x^n$ for equation $(1+x^2)y''+2xy'=0$, valid for large enough $n$.
- $$a_{n+2}=-\frac{2x n}{(1+x^2)(n+2)(n+1)}a_{n}$$
- $$a_{n+2}=-\frac{n}{n+2}a_n$$
- $$a_{n+2}=\frac{(n+1)n}{(n+2)(n+1)} a_{n+1}-\frac{(n-1)}{(n+2)(n+1)}a_n$$
- $$a_{n+2}=-\frac{(n-2)(n-3)}{(n+2)(n+1)}a_n$$
- None of the above
Answer: (B)
Solution Seeking $y=\sum_{n=0}^\infty a_n x^n$. Recall that $$y'=\sum_{n=0}^\infty na_nx^{n-1}\mbox{ so } xy'=\sum_{n=0}^\infty na_nx^{n}$$ $$y''=\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n} \mbox{ but also} $$ $$y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}= \mbox{ so } x^2y''=\sum_{n=2}^\infty n(n-1)a_nx^{n}=\sum_{n=0}^\infty n(n-1)a_nx^{n}$$ Putting the series into the equation (with $(1+x^2)y''=y''+x^2y''$ we get \begin{multline*} \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=0}^\infty n(n-1)a_nx^{n}+2\sum_{n=0}^\infty na_nx^{n}\\= \sum_{n=0}^\infty \left( (n+2)(n+1)a_{n+2} + n(n-1)a_n +2 na_n\right)x^n \\ =\sum_{n=0}^\infty \left((n+2)(n+1)a_{n+2} + n(n+1)a_n \right)x^n =\sum_{n=0}^\infty (n+1) \left((n+2)a_{n+2} + na_n \right)x^n=0 \end{multline*} So we get a recursion $a_{n+2}=-\frac{n}{n+2}a_n$.
Where did $a_0$ disappear? Lets solve this question more completely on the PDF-Board.
Determine the recurrence for the coefficients of the power series solution $y=\sum_{n=0}^\infty a_n x^n$ for equation $(x^2+1)y'' - 4 x y'+6 y=0$, valid for large enough $n$.
- $$a_{n+2}=-\frac{(n-2)(n-3)}{(n+2)(n+1)}a_n$$
- $$a_{n+2}=\frac{4n-6}{(1+x^2)(n+2)(n+1)}a_n$$
- $$a_{n+2}=-\frac{n}{n+2}a_n$$
- $$a_{n+2}=\frac{(n+1)n}{(n+2)(n+1)} a_{n+1}-\frac{(n-1)}{(n+2)(n+1)}a_n$$
- None of the above
Answer: (A)
Solution This is Example 4 in Paul's 6.3
Determine the recurrence for the coefficients of the power series solution $y=\sum_{n=0}^\infty a_n x^n$ for equation $y''+x y=0$, valid for large enough $n$.
- $$a_{n+2}=-\frac{(n-2)(n-3)}{(n+2)(n+1)}a_n$$
- $$a_{n+2}=-\frac{nx}{(n+2)(n+1)}a_n$$
- $$a_{n+2}=-\frac{n}{n+2}a_n$$
- $$a_{n+2}=\frac{(n+1)n}{(n+2)(n+1)} a_{n+1}-\frac{(n-1)}{(n+2)(n+1)}a_n$$
- None of the above
Answer: (E)
Solution $$L[y]=y''+x y'=\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=1}^\infty a_{n-1} x^n$$ So for $n\geq 1$ we have $a_{n+2}=-\frac{a_{n-1}}{(n+2)(n+1)}$, which is none of the choices offered.
What is $a_{2n}$ for $n\geq 1$ as a function of $n,a_0,a_1$, if $a_{n+2}=-\frac{n}{n+2}a_n$
- $$a_{2n}=\frac{a_0}{n!}$$
- $$a_{2n}=(-1)^n \frac{a_0}{n!}$$
- $$a_{2n}=0$$
- $$a_{2n}=(-1)^n\frac{a_0}{n}$$
- None of the above
Answer: C.
Solution
What is $a_{2n}$ for $n\geq 1$ as a function of $n,a_0,a_1$ , if $a_{n+2}=-\frac{n+1}{n+3}a_n$
- $$a_{2n}=(-1)^n\ a_0 $$
- $$a_{2n}=(-1)^n \frac{a_0}{n!}$$
- $$a_{2n}=0$$
- $$a_{2n}=(-1)^n\frac{a_0}{2n+1}$$
- None of the above
Answer: D.
Solution
What is $a_{2n+1}$ for $n\geq 1$ as a function of $n,a_0,a_1$ if $a_{n+2}=-\frac{n}{n+2}a_n$
- $$a_{2n+1}=(-1)^n\frac{a_1}{2n+1}$$
- $$a_{2n+1}=(-1)^n \frac{a_1}{n!}$$
- $$a_{2n+1}=0$$
- $$a_{2n+1}=(-1)^n\frac{a_1}{n}$$
- None of the above
Answer: A.
Solution
Not availableWhat value of $C_1$ will she get for the solution of the initial value problem?
Answer:
Solution
If $y(x)=\sum_{n=0}^\infty a_n x^n$ with $a_{2n}=0$ and $a_{2n+1}=\frac{(-1)^n}{n!}$ then
- $y=e^{-x}$
- $y=x e^{-x}$
- $y=x e^{-x^2}$
- $y=x^2 e^{-x}$
- None of the above
Answer: (C)
Solution $$y= \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{2n+1}=x \sum_{n=0}^\infty \frac{(-x^2)^n}{n!} $$
If $y(x)=\sum_{n=0}^\infty a_n x^n$ with $a_{2n}=0$ and $a_{2n+1}=(-1)^n$ then
- $y=\frac{1}{1+x}$
- $y=\frac{x}{1+x^2}$
- $y=\frac{x^2}{1+x}$
- $y=\frac{x^2}{1+x^2}$
- None of the above
Answer: B
Solution $$y= \sum_{n=0}^\infty (-1)^nx^{2n+1}=x \sum_{n=0}^\infty (-x^2)^n $$
If $y(x)=\sum_{n=0}^\infty a_n x^n$ with $a_{2n}=\frac{(-1)^n}{n!}$ and $a_{2n+1}=0$ then
- $y=e^{-x}$
- $y=x e^{-x}$
- $y=x e^{-x^2}$
- $y=x^2 e^{-x}$
- None of the above
Answer: (E)
Solution $$y= \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{2n}= \sum_{n=0}^\infty \frac{(-x^2)^n}{n!} =e^{-x^2} $$
If $y(x)=\sum_{n=0}^\infty a_n x^n$ with $a_{3n}=0$ , $a_{3n+1}=(-1)^n$ and $a_{3n+2}=0$ yhen
- $y=\frac{1}{1+x^3}$
- $y=\frac{x}{(1+x)^3}$
- $y=\frac{x^3}{1+x}$
- $y=\frac{x^2}{1+x^3}$
- None of the above
Answer: (E)
Solution $$y= \sum_{n=0}^\infty (-1)^nx^{3n+1}=x \sum_{n=0}^\infty (-x^3)^n =\frac{x}{1+x^3}$$
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 $