(I) First-order Equations of Special Form
A differential equation \(\frac{dy}{dx} = f(x, y)\) takes an especially simple form if the right-hand side function \(f\) doesn't actually involve the dependent variable \(y\):
Integrate both sides to get the general solution:
If \(G'(x) = f(x)\), then \(y(x) = G(x) + C\) is the general solution. Geometrically, \(C\) is the vertical distance between two curves:
EX: \(y(x) = G(x)\) and \(Y(x) = G(x) + C\)
\(\frac{dy}{dx} = (x-2)^2, \quad y(2) = 1\)
\(y(x) = \int (x-2)^2 dx = \frac{(x-2)^3}{3} + C\)
\(y(2) = 1 \implies 0 + C = 1 \implies C = 1\)
Solution: \(y(x) = \frac{(x-2)^3}{3} + 1\)
Initial Value Problem
Substitute \(x = x_0\) and \(y = y_0\) into \(y(x) = G(x) + C\): \(C = y_0 - G(x_0)\)
We get the Particular solution: \(y(x) = G(x) + y_0 - G(x_0)\)
(II) Second-order Equations of Special Form
\(g(x)\) doesn't depend on \(\frac{dy}{dx}\) or \(y\). We integrate twice:
- \(\frac{dy}{dx} = \int g(x)dx = G(x) + C_1\)
- \(y(x) = \int (G(x) + C_1)dx = \int G(x)dx + C_1x + C_2\)
Where \(C_1\) and \(C_2\) are arbitrary constants.
(III) Application: Velocity and Acceleration
Case 1: Horizontal Motion
- Position function: \(x(t)\)
- Velocity function: \(v(t) = x'(t)\)
- Acceleration function: \(a(t) = v'(t) = x''(t)\)
Find position function \(x(t)\) of a moving particle with \(a(t) = 2t + 1, \quad v(0) = -7, \quad x(0) = 4\).
\(v(t) = \int a(t)dt = \int(2t+1)dt = t^2 + t + C_1\)
\(v(0) = C_1 = -7 \implies v(t) = t^2 + t - 7\)
\(x(t) = \int v dt = \int(t^2 + t - 7)dt = \frac{t^3}{3} + \frac{t^2}{2} - 7t + C_2\)
\(x(0) = C_2 = 4 \implies x(t) = \frac{t^3}{3} + \frac{t^2}{2} - 7t + 4\)
I.V.P. for 2nd-order D. E.
Case 2: Vertical Motion with Gravitational Acceleration
Using the y-axis for position (\(y=0\) is ground level):
- \(\uparrow\) Pos. velocity; \(\downarrow\) Neg. velocity
- Acceleration: \(a = -g \approx -32 \, \text{ft/s}^2 \approx -9.8 \, \text{m/s}^2\)
Suppose a ball is thrown upward from ground (\(y_0=0\)) with initial velocity \(v_0 = 96 \, \text{ft/s}\). Find the maximum height.
\(v'(t) = -g = -32 \implies v(t) = -32t + 96\)
\(v(t) = 0 \implies t = 3 \, \text{s}\)
\(y(t) = \int(-32t + 96)dt = -16t^2 + 96t + C_2 \quad (y(0) = 0 \implies C_2 = 0)\)
Max Height: \(y(3) = -16 \cdot 3^2 + 96 \cdot 3 = 144 \, \text{ft}\)
At noon, a car starts from rest at Point A toward Point C 35 miles away. Arrives at C with velocity 60 mi/h. At what time does it arrive?
\(v_t = at + v_0 = at = 60 \, \text{mi/h}\)
\(x_t = \frac{1}{2}at^2 + v_0t + x_0 = \frac{1}{2}at^2 = 35\)
\(t = \frac{7}{6} \, \text{h} = 70 \, \text{min}\)
Result: Arrives at 1:10 pm.