MATH 315 FALL 2006 TEST 1 KEY

1. (15 pts.) Solve the differential equation $y' + 2y = t + 3$, with $y(0)=1$.
   Answer: linear with integrating factor $\mu=e^{\int 2dt}=e^{2t}$. Then $(e^{2t}y)' = (t+3)e^{2t}$, so
   $e^{2t}y = \int te^{2t}dt+\int 3e^{2t}dt= te^{2t}/2- e^{2t}/4+3e^{2t}/2 + C$, $y(t) = 5/4 + t/2 + C/e^{2t}$.
   Using $y(0)=5/4 + C= 1$, $C = -1/4$. Final solution: $y(t) = 5/4 + t/2 - e^{-2t}/4$.

2. (15 pts.) Solve the differential equation $\frac{dy}{dx} = \frac{3x-2y}{2x+3y}$, with $y(2)=1$.
   Answer: $(3x-2y)dx+(-2x-3y)dy=0$, so exact with $M_y=N_x=-2$.
   Then $\Psi(x,y)=\int(3x-2y)dx+\int(-2x-3y-\int(-2)dx)dy=3x^2/2-2xy-3y^2/2=C$.
   Using $y(2)=1$, $C = 12/2-2(2)-3/2=1/2$. Final solution: $3x^2 - 4xy -3y^2 = 1$.

3. (15 pts.) Consider the differential equation $\frac{dy}{dt} = t-2y$, with $y(1)=1$.
   1. Draw a rough sketch of the direction field for the equation, plotting direction field arrows for $(t,y)$ values (0,1/2), (0,1), (1/2,1/2), (1/2,1), (1,1/2), (1,1), (3/2,1/2), and (3/2,1).

      
      
      

   2. Write down the general formula used to calculate with the Euler Method.
      Answer: $y_{n+1}=y_n+hf(t_n,y_n)$.

   3. Use the Euler formula to compute an approximate value for y(2), using a step size $h=1/2$ (you will need to take 2 steps).
      Answer: using $f(t,y) = t - 2y$
      first step: $y_0=1$, $t_0=1$, so $f(t_0,y_0)=1-2(1)=-1$, $y_1 = 1 + \frac{1}{2}(-1) = {\bf 1/2}$;
      second step: $y_1=1/2$, $t_1=3/2$, $f(t_1,y_1)=3/2-2(1/2)=1/2$, $y_2 = 1/2 + \frac{1}{2}(1/2) = {\bf 3/4}$.

4. (7 pts.) Consider the differential equation $\frac{dy}{dx} = ((x+1)y)^\frac{1}{2}$. Determine the region in the $x$-$y$ plane where there is a unique solution near each initial point in that region.
   Answer: using $y' = ((x+1)y)^\frac{1}{2}=f(x,y)$, $f_y=((x+1)/y)^\frac{1}{2}/2$, so we need $y \neq 0$ and $(x+1)y >0$ for both $f$ and $f_y$ to exist.
   The $x$-$y$ plane region is given by $x\geq -1$ and $y>0$ or $x\leq -1$ and $y<0$.

5. (18 pts.) Consider the differential equation $\frac{dy}{dt} = -(y-1)(y-3)$.
   1. Determine and classify the equilibrium (critical) points for the equation.
      Answer: $f(y)=-(y-1)(y-3)=0$ when $y=1$ and $y=3$; $f'(y)=-2y+4$.
      $f'(1)=2$ and $f'(3)=-2$, so $y=1$ is unstable and $y=3$ is stable.

   2. Determine the solution for the equation if y(0) = 2.
      Answer: $dy/((y-1)(y-3))=-dt$, so separable.
      Now $1/((y-1)(y-3))= (1/2)(1/(y-1) - 1/(y-3))$, so
      $2\int 1/((y-1)(y-3))dy=\int (1/(y-3)-1/(y-1))dy=-\int 2 dt$, and
      $ln(y-3)-ln(y-1)= -2t+C$, or $(y-3)/(y-1) = ce^{-2t}$.
      Using $y(0)=2$, $(2-3)/(2-1) = -1 = c$. Final solution: $\frac{y-3}{y-1} = -e^{-2t}$; $y(t) = \frac{3 - e^{-2t}}{1-e^{-2t}}.$

6. (10 pts.) Solve the differential equation $x^2y' = x^3 - 3xy$, with $y(-1)=1$.
   Answer: $y'+3y/x = x$, so linear with $\mu=e^{3\int dx/x}=e^{3\ln(x)}=x^3$.
   Then $(x^3y)' = x^4$, $x^3y = x^5/5 + C$, and $y(x) = x^2/5 + C/x^3$.
   Using $y(-1)=1/5 - C= 1$, $C = -4/5$. Final solution: $y(x) = x^2/5 - 4/(5x^3)$.

7. (15 pts.) Consider the differential equation $y'' - y' - 6y = 0$.
   1. Find the general solution for the equation.
      Answer: The characteristic equation is $r^2-r-6=(r-3)(r+2)=0$, so
      $r=3$ or $r=-2$; general solution is $y(t) = c_1e^{3t}+c_2e^{-2t}$.
   2. Determine the Wronskian for fundamental solution set.
      Answer: $W(t) =\left\vert\begin{array}{rr}e^{3t}&e^{-2t} 3e^{3t}&-2e^{-2t}\end{array}\right\vert = -5e^{t}$.
   3. Determine the solution that satisfies initial conditions $y(0)=1$, $y'(0) = 5$.
      Answer: $y(0) = c_1 + c_2=1$ and $y'(0) = 3c_1 - 2c_2=5$,
      so $5c_1 = 7$ and $c_1=7/5$, $c_2=-2/5$. Final solution: $y(t) = 7e^{3t}/5-2e^{-2t}/5$.