To determine the general solution of equation $\alpha x^2 y''+\beta x y'+\gamma y=0$ we seek solution of the form $y=x^r$ and determine the fundamental set of solutions $y_1,y_2$. The general solution $y=C_1 y_1(x)+C_2 y_2(x)$ is
- $y=C_1 x^{r_1}+C_2 x^{r_2}$ (two distinct real roots)
- $y=C_1 x^{r_1}+C_2 x^{r_1}\ln x$ (a double root, i.e. two equal real roots)
- $y=C_1 x^{\lambda}\cos(\mu \ln x)+ C_2 x^{\lambda}\sin(\mu \ln x)$ (two complex roots $r_{1,2}=\lambda\pm i\mu$).
Determine two fundamental solutions of the equation $x^2y''- xy'+ y=0$ for $x>0$.
Answer: $y_1=x$, $y_2=x\ln x$
Solution Substitute $y=x^r$ into the equation. The characteristic equation is $r(r-1)-r+1=0$. This simplifies to $r^2-2r+1=(r-1)^2=0$, so we get double root $r=1$. The general solution of the homogeneous equation is $y=C_1 x+ C_2 x\ln x$
Determine two fundamental solutions of the homogeneous Euler equation $x^2y''-3xy'+4y=0$ for $x>0$.
Answer: $y_1=x^2,y_2=x^2\ln x$
Solution Substitute $y=x^r$ into the equation. The characteristic equation is $r(r-1)-3r+4=0$. This has double root $r=2$. The general solution of the homogeneous equation is $y=C_1 x^2+ C_2 x^2\ln x$
Determine two fundamental solutions of the homogeneous Euler equation $x^2y''+5xy'+4y=0$ for $x>0$.
Answer: $y_1= \frac1{x^{2}}, y_2 \frac{\ln x}{x^2}$
Solution Substitute $y=x^r$ into the equation. The characteristic equation is $r(r-1)+5r+4=0$ with double root $r=-2$. The general solution of the homogeneous equation is $y=C_1 \frac1{x^{2}}+ C_2 \frac{\ln x}{x^2}$
Determine two fundamental solutions of the homogeneous Euler equation $2x^2y''+3xy'-15y=0$ for $x>0$.
Answer: $y_1=x^{5/2},y_2=\frac{1}{x^3}$
Solution
See Paul's 6.4 Example 1
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
$\det\begin{bmatrix} t & t^2 & t^3 \\ 1 & 2t & 3t^2 \\ 0& 2 & 6t \\ \end{bmatrix}=2t^2$
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 $\det\begin{bmatrix} t & t^2 & 1/t \\ 1 & 2t & -1/t^2 \\ 0& 2 & 2/t^3 \\ \end{bmatrix}=\frac{6}{t}$
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 $y=2 e^{-t} + 3 e^{2 t} - 3 \cos(t) - \sin(t)$.
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 $