MATH 315 FALL 2006 TEST 1 REVIEW
Note: the following test was given Fall 1998 when the syllabus was somewhat different. For Fall 2006 review, ignore question 3, and for question 7, ask the question in terms of the Euler formula instead of the Improved Euler formula. Sections 3.1-3 from the text were not covered by this test, but those sections will be covered for Fall 2006 Test 1.

MATH 315 FALL 1998 TEST 1 SOLUTION KEY

1. (15 pts.) Solve the differential equation $y' + 3y = 1 + t$, with $y(0)=2$.

   Answer:
   This is linear so find the integrating factor $\mu = e^{\int 3dt} = e^{3t}$.
   Then $(ye^{3t})' = (1+t)e^{3t}$, so $ye^{3t} = \int(1+t)e^{3t}dt = e^{3t}/3 + te^{3t}/3 - e^{3t}/9 + C$.
   Therefore $y(t) = 2/9 + t/3 + Ce^{-3t}$. Using $y(0)=2$, we have $2/9 + C = 2$, so $C = 16/9$ and
   

   $$y(t) = 2/9 + t/3 + 16e^{-3t}/9.$$
   
   

2. (15 pts.) Solve the differential equation $\frac{dy}{dt} = \frac{y^2-4}{y+1}$, with $y(1)=3$.

   Answer:
   The variables are already separated so use $dt = \frac{y+1}{y^2-4}dy = \frac{y+1}{(y-2)(y+2)}dy = (\frac{A}{y-2}+\frac{B}{y+2})dy$,
   where $A+B = 1$ and $2A-2B = 1$. The solution is $A = 3/4, B = 1/4$.
   We rewrite the equation as $4dt = (\frac{3}{(y-2)}+\frac{1}{(y+2)})dy$, and integrating gives us
   $c + 4t = 3\ln(y-2) + \ln(y+2)$ or $Ce^{4t} = (y-2)^3(y+2)$.
   Using $y(1)=3$, we have $Ce^4 = 5$, so $C = \frac{5}{e^{-4}}$, so the solution is given implicitly by
   

   $$5e^{4(t-1)} = (y-2)^3(y+2).$$
   
   

3. (15 pts.) Solve the differential equation $y^2dx - (x^2+2xy)dy=0$.

   Answer:
   Rewriting the equation as $\frac{dy}{dx} = \frac{y^2}{x^2+2xy}= \frac{(y/x)^2}{1+2(y/x)}$, you can see that it is homogeneous.
   So let $y = xv$ and then the equation becomes $v + x\frac{dv}{dx} = \frac{v^2}{1+2v}$, or $x\frac{dv}{dx} = \frac{v^2}{1+2v}- v = \frac{-v-v^2}{1+2v}$.
   Then $-\frac{dx}{x}=\frac{1+2v}{v+v^2}dv=(\frac{A}{v}+\frac{B}{1+v})dv$, with $A=1$ and $A+B = 2$, so $B=1$.
   Integrating gives us $c-\ln(x) = \ln(v)+\ln(1+v) = \ln(v(1+v))$, so $C/x = v(1+v)= y(x+y)/x^2$ and the final solution is given implicitly by
   

   $$Cx = y(x+y).$$
   
   

4. (15 pts.) Solve the differential equation $\frac{ds}{d\alpha} = \alpha s^2+2s\alpha$, with $s(1)=1$.

   Answer:
   The variables can be separated to give the differential equation $\alpha d\alpha=\frac{1}{s^2+2s}ds$ or
   $\alpha d\alpha=(\frac{A}{s}+\frac{B}{s+2})ds$, where $2A = 1$ and $A+B=0$, so $A = -B = 1/2$. Then $2\alpha d\alpha=(\frac{1}{s}-\frac{1}{s+2})ds$.
   Integrating gives us $c + \alpha^2 = \ln(\frac{s}{2+s})$, or $Ce^{\alpha^2} = \frac{s}{2+s}$, so $C=\frac{1}{3e}$, and the final solution is given implicitly by
   

   $$e^{\alpha^2-1} = 3s/(s+2).$$
   
   

5. (15 pts.) Consider the differential equation $(3xy+y^2)dx + (x^2+xy)dy=0$.

   a)

   Show that the equation is not exact.

   Answer:
   Using $M(x,y) = 3xy+y^2, N(x,y) = x^2+xy$, you can see $M_y = 3x +2y\neq N_x = 2x+y$.

   b)

   Multiply the equation by $x$ and show that the new equation is exact.

   Answer:
   Using the new $M(x,y) = 3x^2y+xy^2, N(x,y) = x^3+x^2y$, you can see
   $M_y = 3x^2 +2xy = N_x = 3x^2+2xy$.

   c)

   Solve the equation for the solution that satisfies $y(1) = 1$.

   Answer:
   You have $\Psi(x,y) = \int(3x^2y+xy^2)dx + \int(x^3+x^2y-\int(M_y)dx)dy$, so
   $\Psi(x,y)= x^3y+\frac{x^2y^2}{2} + \int(x^3+x^2y-(x^3+x^2y))dy = x^3y+\frac{x^2y^2}{2}= C$,
   so $C = 3/2$ and the final solution is given implicitly by
   

   $$x^3y+x^2y^2/2 = 3/2.$$

   
   

6. (10 pts.) Consider the differential equation $\frac{dy}{dt} = (y+2t)^\frac{1}{2}$. Determine the region in the $ty$ plane where a unique solution should exist.

   Answer:
   Using $f(t,y)=(y+2t)^\frac{1}{2}$, and $f_y = \frac{1}{2(y+2t)^\frac{1}{2}}$, you need to determine where $f$ and/or $f_y$ do not exist. This happens if $y + 2t \leq 0$, so the region is determined by
   

   $$y > -2t.$$
   
   

7. (15 pts.) Consider the differential equation $\frac{dy}{dt} = y+t$, with $y(0)=1$.

   a)

   Write down the general formula used to calculate with the Improved Euler Method.

   Answer:
   $y_{n+1} = y_n + \frac{h}{2}(\ f(t_n,y_n) + f(t_{n+1},y_n+hf(t_n,y_n))\ )$.

   b)

   Use the Improved Euler formula to compute an approximate solution to the equation at $t=1$, using a step size $h=1/2$ (you will need to take two steps).
   

   Answer:
   If you use $f(t_n,y_n) = t_n + y_n$ then the formula becomes
   $y_{n+1} = y_n + \frac{h}{2}(t_n+y_n + t_n+h+ y_n+ht_n+hy_n) = y_n + h(t_n+y_n) + h^2(1+ t_n+y_n)/2$.
   Starting with $t_0=0$, $y_0=1$, $t_1=1/2$, and using $h=1/2$
   $y_1 = 1 + 1/2 + (2)/8 = 7/4$, and using $t_2 = 1$,
   $y_2=7/4+(1/2+7/4)/2+(3/2+7/4)/8=7/4+9/8+13/32= 105/32$. Therefore
   

   $$y(1) \simeq 105/32.$$