Math 308 Week in Review 1 (with Solutions)

Let \( u(x) = e^{2x} \). Then the first derivative is \( u'(x) = 2e^{2x} \) and the second is \( u''(x) = 4e^{2x} \).

Substituting into the equation gives:
\( 4e^{2x} - 4(2e^{2x}) + 4(e^{2x}) = 4e^{2x} - 8e^{2x} + 4e^{2x} = 0 \).

The equation holds true for all \( x \), so yes, \( e^{2x} \) is a solution.

Let \( h(x) = \sin(3x) \). The derivatives are:
\( h'(x) = 3\cos(3x) \),
\( h''(x) = -9\sin(3x) \),
\( h'''(x) = -27\cos(3x) \),
\( h^{(4)}(x) = 81\sin(3x) \).

Substituting into the left side:
\( 81\sin(3x) - (-9\sin(3x)) = 81\sin(3x) + 9\sin(3x) = 90\sin(3x) \).

This is not equal to \( 8\sin(3x) \), so \( \sin(3x) \) is not a solution.

Let \( p(t) = \cos(2t) \). Then \( p'(t) = -2\sin(2t) \) and \( p''(t) = -4\cos(2t) \).

Substituting into the differential equation:
\( -4\cos(2t) + (-2\sin(2t)) + 4\cos(2t) = -2\sin(2t) \),
which is not equal to \( 2\sin(2t) \).

Although the initial condition \( p(0) = \cos(0) = 1 \) is satisfied, the function does not satisfy the differential equation itself. Therefore, \( \cos(2t) \) is not a solution to the IVP.

Assume \( w(t) = e^{rt} \). Then \( w'(t) = r e^{rt} \) and \( w''(t) = r^2 e^{rt} \).

Substituting into the equation:
\( r^2 e^{rt} - 5 r e^{rt} + 6 e^{rt} = 0 \).
Dividing through by \( e^{rt} \) (which is never zero) gives the characteristic equation:
\( r^2 - 5r + 6 = 0 \).

Factoring yields \( (r-2)(r-3) = 0 \), so the roots are \( r = 2 \) and \( r = 3 \).

Let \( v(x) = \cos(kx) \). The second derivative is \( v''(x) = -k^2 \cos(kx) \).

Substituting into the equation:
\( \cos(kx) + 9(-k^2 \cos(kx)) = (1 - 9k^2) \cos(kx) = 0 \).

For this to hold for all \( x \), we must have \( 1 - 9k^2 = 0 \), so \( k^2 = \dfrac{1}{9} \), and therefore \( k = \pm \dfrac{1}{3} \).

Assume \( y(t) = t^m \) for \( t > 0 \). Then \( y'(t) = m t^{m-1} \) and \( y''(t) = m(m-1) t^{m-2} \).

Substituting into the equation:
\( t^2 \cdot m(m-1) t^{m-2} - 3t \cdot m t^{m-1} + 3 t^m = m(m-1) t^m - 3m t^m + 3 t^m = [m(m-1) - 3m + 3] t^m = 0 \).

This simplifies to the characteristic equation \( m^2 - 4m + 3 = 0 \), or \( (m-1)(m-3) = 0 \).
Thus the solutions are \( m = 1 \) and \( m = 3 \).

(a) This is a PDE because it involves partial derivatives with respect to two independent variables, \( s \) and \( t \). The highest-order derivative is first order, so it is order 1.

(b) This is an ODE (only one independent variable, \( x \)). The highest derivative is second order, so the equation is of order 2.

(c) This is an ODE (only \( z \) is the independent variable). The highest derivative is third order, so the equation is of order 3.

(d) This is a PDE (partial derivatives with respect to \( x \) and \( y \)). The highest-order derivatives are second order, so it is order 2.

(a) Nonlinear — the term \( u'' u \) is a product of the function and its second derivative.

(b) Nonlinear — the term \( 1/h'(t) \) involves the reciprocal of a derivative.

(c) Linear and homogeneous — after moving all terms to one side we get \( P'''(x) + \sin(x) P'(x) - (x^2 + \cos(x)) P(x) = 0 \), which is linear with no forcing term.

(d) Nonlinear — the term \( x^3 w^2 \) is quadratic in \( w \).

(e) Linear and nonhomogeneous — the equation can be written as \( v^{(6)} + \sin(s) v''' = \cos(s) - s^2 \), which is linear in \( v \) and its derivatives, but the right-hand side is nonzero.

Rewrite the equation as \( u' = u(u - 4) \). Equilibrium solutions occur where \( u' = 0 \), so \( u = 0 \) and \( u = 4 \).

The sign of \( u' \) determines the direction of the slopes:
• When \( u < 0 \), \( u' > 0 \) → solutions increase.
• When \( 0 < u < 4 \), \( u' < 0 \) → solutions decrease.
• When \( u > 4 \), \( u' > 0 \) → solutions increase.

Thus \( u = 0 \) is a stable equilibrium (solutions approach it from both sides within the interval), while \( u = 4 \) is unstable (solutions move away from it).

Long-time behavior as \( t \to \infty \):
• If \( c < 4 \) (including negative values), \( u(t) \to 0 \).
• If \( c = 4 \), \( u(t) = 4 \) for all time.
• If \( c > 4 \), \( u(t) \to +\infty \).

Part 1 – Matching direction fields:
Figure (a) → (b) \( y' = y(2-y) \) (stable node at 2, unstable at 0).
Figure (b) → (d) \( y' = -y(2-y) = y(y-2) \) (stable at 0, unstable at 2).
Figure (c) → (a) \( y' = y(y-2)^2 \) (unstable at 0, semi-stable at 2).
Figure (d) → (c) \( y' = y + 2 \) (unstable equilibrium at -2).

Part 2 – Long-term behavior as \( t \to \infty \):
Figure (a): Solutions approach 2 from below and above if starting near it; go to \(-\infty\) if starting below 0.
Figure (b): Solutions approach 0 if starting below 2; diverge to \(+\infty\) if starting above 2.
Figure (c): Solutions go to \(+\infty\) if starting above 0; go to \(-\infty\) if starting below 0.
Figure (d): Solutions go to \(+\infty\) if starting above -2; go to \(-\infty\) if starting below -2.

(a) Separable (can write as \( \cos(v) \, dv = \sin(t) \, dt \)); nonlinear.

(b) Neither clearly separable nor linear in standard form.

(c) Separable (rewrite as \( w^3 \, dw = 5 x^{-1} \, dx \)); nonlinear.

(d) Both separable and linear (first-order linear with constant coefficients).

(e) Linear (standard first-order form \( h' - 6h = -2t \)); not separable.

(f) Linear (can be written as \( v' = t^2 v - 4 \)); not separable.

(g) Linear (standard form \( q' - (t+3)q = t+3 \)); separable after solving for q if desired.

Rewrite the equation as \( u' - 3u = -12 \). This is a first-order linear equation, but we can also solve it by separation of variables.

Separate: \( \dfrac{du}{u - 4} = 3 \, dt \).
Integrate both sides: \( \ln|u - 4| = 3t + C \).
Exponentiate: \( u - 4 = A e^{3t} \) where \( A = \pm e^C \).
So \( u(t) = 4 + A e^{3t} \).

Apply the initial condition \( u(0) = 5 \):
\( 5 = 4 + A \Rightarrow A = 1 \).

Therefore the solution is \( u(t) = 4 + e^{3t} \).

This is separable. Rewrite as:
\( h^2 \, dh = (x^3 + 5) \, dx \).

Integrate both sides:
\( \int h^2 \, dh = \int (x^3 + 5) \, dx \),
\( \dfrac{h^3}{3} = \dfrac{x^4}{4} + 5x + C \).

Multiply through by 3:
\( h^3 = \dfrac{3}{4} x^4 + 15x + K \), where \( K = 3C \).

Therefore the general solution is:
\( h(x) = \left( \dfrac{3}{4} x^4 + 15x + K \right)^{1/3} \).

Separate variables:
\( e^{h} \, dh = (7 - 2x) \, dx \).

Integrate both sides:
\( e^{h} = 7x - x^2 + C \).
So \( h(x) = \ln(7x - x^2 + C) \).

Apply the initial condition \( h(3) = 2 \):
\( 2 = \ln(21 - 9 + C) \Rightarrow e^2 = 12 + C \Rightarrow C = e^2 - 12 \).

The solution is therefore:
\( h(x) = \ln(7x - x^2 + e^2 - 12) \).

The expression inside the logarithm must be positive:
\( 7x - x^2 + e^2 - 12 > 0 \).
Rearrange: \( -x^2 + 7x + (e^2 - 12) > 0 \).
This quadratic inequality holds between the two roots (approximately \( 0.7361 < x < 6.2639 \) ). Thus the solution is defined on that open interval.

(a) Model and solution
Let \( u(t) \) be the temperature of the coffee at time \( t \) (in minutes). By Newton's Law of Cooling, the rate of change is proportional to the difference between the coffee temperature and the room temperature:
\[ u' = -k (u - 20), \quad u(0) = 95. \]

Make the substitution \( y = u - 20 \). Then \( y' = u' \) and the equation becomes:
\[ y' = -k y, \quad y(0) = 95 - 20 = 75. \]

This is a standard separable equation of the form \( y' = a y \) with \( a = -k \). The general solution is:
\[ y(t) = C e^{a t} = C e^{-k t}. \]

Apply the initial condition:
\[ y(0) = C = 75 \quad \Rightarrow \quad y(t) = 75 e^{-k t}. \]

Back-substitute to obtain the solution for \( u \):
\[ u(t) = 20 + y(t) = 20 + 75 e^{-k t}. \]

Use the given data at \( t = 5 \): \( u(5) = 70 \)
\[ 70 = 20 + 75 e^{-5k} \quad \Rightarrow \quad 50 = 75 e^{-5k} \quad \Rightarrow \quad e^{-5k} = \dfrac{50}{75} = \dfrac{2}{3}. \]
Taking the natural log:
\[ -5k = \ln\left(\dfrac{2}{3}\right) \quad \Rightarrow \quad k = -\dfrac{1}{5} \ln\left(\dfrac{2}{3}\right) = \dfrac{1}{5} \ln\left(\dfrac{3}{2}\right). \]

(b) Temperature at t = 10 minutes
\[ u(10) = 20 + 75 \left( \dfrac{2}{3} \right)^2 = 20 + 75 \cdot \dfrac{4}{9} = 20 + \dfrac{300}{9} = 20 + 33.\overline{3} \approx 53.33^\circ\text{C}. \] (c) Time to reach 40°C
Set \( u(t) = 40 \):
\[ 40 = 20 + 75 e^{-k t} \quad \Rightarrow \quad 20 = 75 e^{-k t} \quad \Rightarrow \quad e^{-k t} = \dfrac{20}{75} = \dfrac{4}{15}. \]
\[ -k t = \ln\left(\dfrac{4}{15}\right) \quad \Rightarrow \quad t = -\dfrac{1}{k} \ln\left(\dfrac{4}{15}\right) = \dfrac{1}{k} \ln\left(\dfrac{15}{4}\right). \]

Substitute \( k = \dfrac{1}{5} \ln\left(\dfrac{3}{2}\right) \):
\[ t = 5 \cdot \dfrac{\ln(15/4)}{\ln(3/2)} \approx 16.3 \text{ minutes}. \]