- The general solution of the constant coefficient non-homogeneous equation $a_0y^{(n)}+a_1 y^{(n-1)}+\dots+a_{n-1}y'+a_n y=g(t)$ is $$y(t)=\left(C_1 y_1 +C_2 y_2+\dots+C_ny_n\right)+y_*$$ where $C_1 y_1 +C_2 y_2+\dots+C_ny_n$ is the general solution of the corresponding homogeneous equation $a_0y^{(n)}+a_1 y^{(n-1)}+\dots+a_{n-1}y'+a_n y=0$, and $y_*$ is a particular solution of the non-homogeneous equation.
- Method of undetermined coefficients:
$g(t)$ $y_*$ $1$ $t^{s}A$ $t$ $t^{s}(A+Bt)$ $ e^{\alpha t}$ $t^{s}A e^{\alpha t}$ $ t e^{\alpha t}$ $t^{s}(A +B t)e^{\alpha t}$ $ \cos(\alpha t)$ $t^{s}(A\cos (\alpha t) +B \sin(\alpha t))$ - Variation of parameters for third-order equation $L[y]=g(t)$ with fundamental set $y_1,y_2,y_3$ and $t_0=0$: $$y_*=y_1(t)\int_0^t \frac{W_1}{W} g(s)ds+y_2(t)\int_0^t \frac{W_2}{W} g(s)ds+y_3(t)\int_0^t \frac{W_3}{W}g(s) ds$$
According to WolframAlpha, the general solution of a homogeneous equation $L[y]=0$ with constant coefficients is $y=C_1e^{-t}+C_2 e^{t}+C_3 t e^{-t}+C_4 t e^t$ Which of the following is the right choice for a solution of the equation $$ L[y]=e^{-t} $$ by the method of undetermined coefficients?
- $y_*=A e^{-t}$
- $y_*=A t e^{-t}$
- $y_*=A t^2 e^{-t}$
- $y_*=A t^3 e^{-t}$
- none of these
Answer: C
Solution
From $y=C_1e^{-t}+C_2 e^{t}+C_3 t e^{-t}+C_4 t e^t$ we see that there are two double roots $r=\pm1$. The multiplicity of root $r=-1$ is $s=2$. So (C) is the right guess. We can also recover the equation: $(r-1)^2(r+1)^2=r^4-2r^2+1$ so the equation to solve was $y^{(4)}-2y''+y=e^{-t}$.
According to WolframAlpha, the general solution of a homogeneous equation $L[y]=0$ with constant coefficients is $y=C_1\cos t+C_2 t\cos t+C_3\sin t+C_4 t\sin t$ Which of the following is the right choice for a solution of the equation $$ L[y]=e^{-t} $$ by the method of undetermined coefficients?
- $y_*=A e^{-t}$
- $y_*=A t e^{-t}$
- $y_*=A t^2 e^{-t}$
- $y_*=A t^3 e^{-t}$
- none of these
Answer: A
Solution From the general solution we see that there are two double roots $r=\pm i$. The multiplicity of root $r=-1$ is $s=0$ (not a root). We can also recover the equation: $(r-i)^2(r+i)^2=r^4+2r^2+1$ so the equation to solve was $y^{(4)}+2y''+y=e^{-t}$.
According to WolframAlpha, the general solution of a homogeneous equation $L[y]=0$ with constant coefficients is $y=C_1\cos t+C_2 t\cos t+C_3\sin t+C_4 t\sin t$ Which of the following is the right choice for a solution of the equation $$ L[y]=\sin t $$ by the method of undetermined coefficients?
- $y_*=A \sin t+B\cos t$
- $y_*=A t \sin t+B t \cos t$
- $y_*=A t^2 \sin t+B t^2 \cos t$
- $y_*=A t^3 \sin t+B t^3 \cos t$
- none of these
Answer: C
Solution
From the general solution we see that there are two double roots $r=\pm i$.
Which of the following is the right choice for a solution of the equation $$ y''''+2y''+y=\cos t $$ by the method of undetermined coefficients?
- $y_*=A \sin t+B\cos t$
- $y_*=A t \sin t+B t \cos t$
- $y_*=A t^2 \sin t+B t^2 \cos t$
- $y_*=A t^3 \sin t+B t^3 \cos t$
- none of these
Answer: C
Solution The general solution of the homogeneous equation is $y=C_1\cos t+C_2 t\cos t+C_3\sin t+C_4 t\sin t$. The multiplicity of roots $r=\pm i$ is $s=2$.
To solve the initial value problem $y''' - 3 y''- y'+ 3 y=18 e^{2t},\; y(0)=2,y'(0)=0$ a student seeks a particular solution of the form $y_*=A t^2 e^{2t}$.What value of $A$ will she get?
Answer: none
Solution
There is no particular solution of the form $At^2 e^{2t}$. The roots of the characteristic equation are $r_1=-1$, $r_2=1$, $r_3=3$. One can solve this problem by hand, or you can read this answer from WolframAlpha
According to WolframAlpha, the solution of $y'''+y''-4y'-4y=g(t)$ is $$y(t)=e^{-2 t} \left(\int _1^t\frac{1}{4} e^{2 \xi} g(\xi )d\xi+e^t \int _1^t-\frac{1}{3} e^{\zeta} g(\zeta)d\zeta+e^{4 t} \int _1^t\frac{1}{12} e^{-2 \upsilon} g(\upsilon)d\upsilon \right)+c_1 e^{-2 t}+c_2 e^{-t}+c_3 e^{2 t}$$What is the particular solution with $t_0=0$ according to the variation of parameters method?
Answer: $e^{-2 t} \int _0^t\frac{1}{4} e^{2 s} g(s)ds+e^{-t} \int _0^t-\frac{1}{3} e^{s} g(s)ds+e^{2 t} \int _0^t\frac{1}{12} e^{-2 s} g(s)ds $
Solution
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 $